The Calculus Course

1. Distance is the accumulation of velocity over time.
2. On a graph of velocity versus time, the total distance is represented by the area between the curve and the time axis.
3. Therefore, this is the area problem.

Answer: The area problem.

1. The slope between two points is a secant line.
2. As the points merge into one, the line no longer crosses the curve but just touches it.
3. A line that touches a curve at a single point is called a tangent line.

Answer: The tangent line.

1. The formula $\text{length} \times \text{width}$ only works for rectangles.
2. A rectangle has a constant height.
3. The curve $f(x) = x^2$ has a height that changes at every single point along the x axis.
4. Because the height is not constant, there is no single 'width' or 'length' that describes the whole shape.

Answer: Because the height of the shape is constantly changing, so it is not a rectangle.

1. Substitute $-4$ for every $x$ in the expression: $f(-4) = (-4)^{2} + 3(-4) - 5$.
2. Calculate the square: $(-4)^{2} = 16$.
3. Calculate the multiplication: $3(-4) = -12$.
4. Combine the terms: $16 - 12 - 5 = 4 - 5 = -1$.

Answer: -1

1. Identify the operation that could be undefined: division by zero.
2. Set the denominator equal to zero to find the restricted value: $x - 3 = 0$.
3. Solve for $x$: $x = 3$.
4. The function is defined for all real numbers except $x = 3$.

Answer: x \neq 3

1. Find the domain by looking at the x-coordinates of the endpoints: the leftmost is $-1$ and the rightmost is $5$.
2. The domain is $[-1, 5]$.
3. Find the range by looking at the y-coordinates of the endpoints: the lowest is $2$ and the highest is $10$.
4. The range is $[2, 10]$.

Answer: Domain: [-1, 5], Range: [2, 10]

1. Calculate the slope $m = \frac{10 - 4}{2 - (-1)} = \frac{6}{3} = 2$.
2. Substitute $m = 2$ and the point $(2, 10)$ into $y = mx + c$.
3. $10 = 2(2) + c \implies 10 = 4 + c \implies c = 6$.
4. Write the final equation.

Answer: $y = 2x + 6$

1. Find $f(0) = -\frac{1}{2}(0) + 3 = 3$.
2. Find $f(4) = -\frac{1}{2}(4) + 3 = -2 + 3 = 1$.
3. Calculate the average rate of change: $\frac{f(4) - f(0)}{4 - 0} = \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2}$.

Answer: $-\frac{1}{2}$

1. Start with the equation $y = mx + c$.
2. Substitute $m = 3$, $x = 4$, and $y = -2$.
3. $-2 = 3(4) + c$.
4. $-2 = 12 + c$.
5. Subtract $12$ from both sides to find $c = -14$.

Answer: $-14$

1. Set the function to zero: $x^2 - 5x + 6 = 0$.
2. Factorise the quadratic: $(x - 2)(x - 3) = 0$.
3. Solve for $x$: $x = 2$ and $x = 3$.
4. Identify the degree as 2 (even) and the leading coefficient as 1 (positive), so both ends point upwards.

Answer: Roots: $x=2, 3$; End behavior: both ends point up.

1. Factor out the common term $x$: $f(x) = x(x^2 - 5x + 6)$.
2. Factorise the quadratic part: $f(x) = x(x - 2)(x - 3)$.
3. Set each factor to zero: $x = 0$, $x = 2$, and $x = 3$.
4. Identify the degree as 3 (odd) and the leading coefficient as 1 (positive), so the graph goes down to the left and up to the right.

Answer: Roots: $x=0, 2, 3$; End behavior: down to the left, up to the right.

1. Factor out the common term $x$: $f(x) = x(x^2 + 1)$.
2. Set the factors to zero: $x = 0$ or $x^2 + 1 = 0$.
3. Analyze $x^2 + 1 = 0$: this implies $x^2 = -1$.
4. Since no real number squared is negative, $x^2 + 1$ has no real roots.

Answer: No, the only real root is $x = 0$.

1. The starting amount is $20$ and the growth factor is $2$, so the function is $f(x) = 20 \times 2^x$.
2. Substitute $x = 5$ into the function: $f(5) = 20 \times 2^5$.
3. Calculate $2^5 = 32$.
4. Multiply $20 \times 32 = 640$.

Answer: 640

1. Continuous growth at $100\%$ is modeled by the base $e$.
2. The function is $f(t) = 1 \times e^t$.
3. Substitute $t = 3$: $f(3) = e^3$.
4. Calculate $2.718 \times 2.718 \times 2.718 \approx 20.079$.

Answer: 20.079

1. Evaluate $g(4) = 4^3 = 4 \times 4 \times 4 = 64$.
2. Evaluate $h(4) = 3^4 = 3 \times 3 \times 3 \times 3 = 81$.
3. Compare the two: $81 > 64$.
4. Find the difference: $81 - 64 = 17$.

Answer: h(x) is larger by 17

1. Take the natural logarithm of both sides: $\ln(3^x) = \ln(20)$
2. Apply the power law to bring $x$ to the front: $x \ln(3) = \ln(20)$
3. Divide both sides by $\ln(3)$ to isolate $x$: $x = \frac{\ln(20)}{\ln(3)}$

Answer: $x = \frac{\ln(20)}{\ln(3)}$

1. Use the product law: $\log_5(125) + \log_5(5^2)$
2. Evaluate $\log_5(125)$ since $5^3 = 125$, so $\log_5(125) = 3$
3. Use the power law for the second term: $2 \log_5(5) = 2 \cdot 1 = 2$
4. Add the results: $3 + 2 = 5$

Answer: 5

1. Take the natural logarithm of both sides: $\ln(e^{2x}) = \ln(7)$
2. Since $\ln$ and $e$ are inverses, $\ln(e^{2x}) = 2x$
3. The equation becomes $2x = \ln(7)$
4. Divide by 2: $x = \frac{\ln(7)}{2}$

Answer: $x = \frac{\ln(7)}{2}$

1. The angle $\frac{5\pi}{6}$ is in the second quadrant because it is between $\frac{\pi}{2}$ and $\pi$.
2. The reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$ (or $30^\circ$).
3. For $\frac{\pi}{6}$, the coordinates are $(\frac{\sqrt{3}}{2}, \frac{1}{2})$.
4. In the second quadrant, $x$ is negative and $y$ is positive, so the point is $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$.
5. Therefore, $\sin(\frac{5\pi}{6}) = \frac{1}{2}$, $\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}$, and $\tan(\frac{5\pi}{6}) = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$.

Answer: \sin(\frac{5\pi}{6}) = \frac{1}{2}, \cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}, \tan(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{3}

1. To convert to radians, multiply by $\frac{\pi}{180}$: $225 \times \frac{\pi}{180} = \frac{225\pi}{180} = \frac{5\pi}{4}$.
2. The angle $\frac{5\pi}{4}$ is in the third quadrant.
3. The reference angle is $\frac{5\pi}{4} - \pi = \frac{\pi}{4}$.
4. The cosine of $\frac{\pi}{4}$ is $\frac{\sqrt{2}}{2}$.
5. In the third quadrant, cosine is negative, so $\cos(225^\circ) = -\frac{\sqrt{2}}{2}$.

Answer: \frac{5\pi}{4} \text{ radians}, \cos(225^\circ) = -\frac{\sqrt{2}}{2}

1. Use the identity $\sin^2(\theta) + \cos^2(\theta) = 1$.
2. Substitute the known value: $(\frac{3}{5})^2 + \cos^2(\theta) = 1 \implies \frac{9}{25} + \cos^2(\theta) = 1$.
3. Solve for $\cos^2(\theta)$: $\cos^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25}$.
4. Take the square root: $\cos(\theta) = \pm \frac{4}{5}$.
5. Since $\theta$ is in the second quadrant, $\cos(\theta)$ must be negative, so $\cos(\theta) = -\frac{4}{5}$.
6. Find tangent: $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{3/5}{-4/5} = -\frac{3}{4}$.

Answer: \cos(\theta) = -\frac{4}{5}, \tan(\theta) = -\frac{3}{4}

1. We look for an angle $\theta$ such that $\cos(\theta) = -\frac{\sqrt{2}}{2}$ within the range $[0, \pi]$.
2. The reference angle for $\frac{\sqrt{2}}{2}$ is $\frac{\pi}{4}$.
3. Since the cosine is negative, the angle must be in the second quadrant.
4. The angle is $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Answer: \frac{3\pi}{4}

1. We look for an angle $\theta$ such that $\tan(\theta) = -1$ within the range $(-\frac{\pi}{2}, \frac{\pi}{2})$.
2. We know $\tan(\frac{\pi}{4}) = 1$.
3. Since tangent is an odd function, $\tan(-\frac{\pi}{4}) = -1$.
4. The angle $-\frac{\pi}{4}$ is within the required range.

Answer: -\frac{\pi}{4}

1. First, evaluate the inner part: $\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$.
2. Now we need to find $\arcsin(\frac{\sqrt{3}}{2})$.
3. We look for an angle $\theta$ in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin(\theta) = \frac{\sqrt{3}}{2}$.
4. The angle $\frac{\pi}{3}$ satisfies this and is within the range.

Answer: \frac{\pi}{3}

1. Identify the inner function $g(x) = 3x - 2$.
2. Substitute $g(x)$ into $f(x)$: $f(g(x)) = (3x - 2)^2$.
3. Expand the square: $(3x)^2 - 2(3x)(2) + 2^2$.
4. Simplify the terms: $9x^2 - 12x + 4$.

Answer: $f(g(x)) = 9x^2 - 12x + 4$

1. The multiplier $3$ outside the function indicates a vertical stretch by a factor of $3$.
2. The term $-\frac{\pi}{4}$ inside the function indicates a horizontal shift to the right by $\frac{\pi}{4}$ units.
3. The constant $+1$ at the end indicates a vertical shift upwards by $1$ unit.

Answer: Vertical stretch by 3, horizontal shift right by $\frac{\pi}{4}$, and vertical shift up by 1.

1. The composite function is $f(g(x)) = \frac{1}{\sin(x)}$.
2. The inner function $g(x) = \sin(x)$ is defined for all real numbers.
3. The outer function $f(u) = \frac{1}{u}$ is undefined where $u = 0$. Thus, we need $\sin(x) \neq 0$.
4. In the interval $[0, \pi]$, $\sin(x) = 0$ at $x = 0$ and $x = \pi$.

Answer: $x \neq 0, \pi$

1. Add $7$ to both sides to get $3x \ge 21$.
2. Divide both sides by $3$ to get $x \ge 7$.
3. Since $x$ is greater than or equal to $7$, the interval starts at $7$ and goes to infinity.
4. Use a square bracket for $7$ because it is included.

Answer: [7, \infty)

1. Set up two cases: $2x + 5 = 11$ and $2x + 5 = -11$.
2. For the first case, subtract $5$ to get $2x = 6$, then divide by $2$ to get $x = 3$.
3. For the second case, subtract $5$ to get $2x = -16$, then divide by $2$ to get $x = -8$.
4. Both values satisfy the original equation.

Answer: x = 3, x = -8

1. The distance from $x$ to $3$ must be greater than $2$.
2. This means $x - 3 > 2$ or $x - 3 < -2$.
3. Solving the first part: $x > 5$, which is $(5, \infty)$.
4. Solving the second part: $x < 1$, which is $(-\infty, 1)$.
5. Combine these two disjoint sets using the union symbol.

Answer: (-\infty, 1) \cup (5, \infty)

1. Step 1: Plug in values close to $1$ from the left, such as $0.9$ and $0.99$. $f(0.9) = 0.81 + 2 = 2.81$ and $f(0.99) = 0.9801 + 2 = 2.9801$.
2. Step 2: Plug in values close to $1$ from the right, such as $1.1$ and $1.01$. $f(1.1) = 1.21 + 2 = 3.21$ and $f(1.01) = 1.0201 + 2 = 3.0201$.
3. Step 3: Observe that both sides are approaching the value $3$.

Answer: 3

1. Step 1: Notice that plugging in $x = 4$ results in $0/0$, which is undefined.
2. Step 2: Factor the numerator as a difference of squares: $x^2 - 16 = (x - 4)(x + 4)$.
3. Step 3: Simplify the function: $f(x) = \frac{(x - 4)(x + 4)}{x - 4} = x + 4$ for $x \neq 4$.
4. Step 4: As $x$ approaches $4$, the expression $x + 4$ approaches $4 + 4 = 8$.

Answer: 8

1. Step 1: Approach $x = 2$ from the left. For all $x < 2$, $f(x) = 5$, so the left-hand limit is $5$.
2. Step 2: Approach $x = 2$ from the right. For all $x \ge 2$, $f(x) = 10$, so the right-hand limit is $10$.
3. Step 3: Compare the two sides. Since $5 \neq 10$, the function does not approach a single value.

Answer: No, the limit does not exist because the left-hand and right-hand limits are different.

1. Factorise the numerator: $x^2 + 5x + 6 = (x + 2)(x + 3)$
2. Rewrite the limit: $\lim_{x \to -2} \frac{(x + 2)(x + 3)}{x + 2}$
3. Cancel the common factor $(x + 2)$ for $x \neq -2$: $\lim_{x \to -2} (x + 3)$
4. Substitute $x = -2$ into the simplified expression: $-2 + 3 = 1$

Answer: 1

1. Split the absolute value into two cases: $3x + 1 > 7$ or $3x + 1 < -7$
2. Solve the first case: $3x > 6 \implies x > 2$
3. Solve the second case: $3x < -8 \implies x < -8/3$
4. Combine the results into interval notation: $(-\infty, -8/3) \cup (2, \infty)$

Answer: (-\infty, -8/3) \cup (2, \infty)

1. Substitute $g(x)$ into $f(x)$: $(f \circ g)(x) = f(2x - 5) = e^{2x - 5}$
2. Substitute $x = 3$ into the composite function: $e^{2(3) - 5}$
3. Simplify the exponent: $e^{6 - 5} = e^1$
4. The final value is $e$

Answer: e

1. Pick values approaching 3 from the left: $x = 2.9, 2.99, 2.999$.
2. Calculate $f(2.9) = \frac{8.41 - 9}{2.9 - 3} = \frac{-0.59}{-0.1} = 5.9$.
3. Calculate $f(2.99) = \frac{8.9401 - 9}{2.99 - 3} = \frac{-0.0599}{-0.01} = 5.99$.
4. Pick values approaching 3 from the right: $x = 3.1, 3.01, 3.001$.
5. Calculate $f(3.1) = \frac{9.61 - 9}{3.1 - 3} = \frac{0.61}{0.1} = 6.1$.
6. Calculate $f(3.01) = \frac{9.0601 - 9}{3.01 - 3} = \frac{0.0601}{0.01} = 6.01$.
7. Observe that both sides approach 6.

Answer: 6

1. Find the left-hand limit: $\lim_{x \to 1^{-}} (x + 5) = 1 + 5 = 6$.
2. Find the right-hand limit: $\lim_{x \to 1^{+}} (2x + 4) = 2(1) + 4 = 6$.
3. Since the left-hand limit equals the right-hand limit, the general limit exists.

Answer: 6

1. Approach 0 from the right: as $x \to 0^{+}$, $1/x$ becomes very large, so $\lim_{x \to 0^{+}} \frac{1}{x} = \infty$.
2. Approach 0 from the left: as $x \to 0^{-}$, $1/x$ becomes very large and negative, so $\lim_{x \to 0^{-}} \frac{1}{x} = -\infty$.
3. Since the left-hand and right-hand limits are not equal (and are not finite), the limit does not exist.

Answer: Does Not Exist

1. Compute the left-hand limit: $\lim_{x \to -1^{-}} x^2 = (-1)^2 = 1$.
2. Compute the right-hand limit: $\lim_{x \to -1^{+}} (-x) = -(-1) = 1$.
3. Since the left-hand limit equals the right-hand limit, the overall limit exists.

Answer: 1

1. As $x$ approaches $2$ from the right, $(x-2)^2$ is a very small positive number, so $1/(x-2)^2$ goes to $\infty$.
2. As $x$ approaches $2$ from the left, $(x-2)^2$ is also a very small positive number, so $1/(x-2)^2$ goes to $\infty$.
3. Because the function is unbounded, it does not approach a finite real number.

Answer: DNE (Unbounded)

1. For $x > -3$, $|x+3| = x+3$, so $f(x) = (x+3)/(x+3) = 1$. The right-hand limit is $1$.
2. For $x < -3$, $|x+3| = -(x+3)$, so $f(x) = -(x+3)/(x+3) = -1$. The left-hand limit is $-1$.
3. Since $1 \neq -1$, the limit does not exist.

Answer: DNE (Jump)

1. Substitute $x = 5$ to check for indeterminate form: $\frac{5^2 - 25}{5 - 5} = \frac{0}{0}$.
2. Factor the numerator as a difference of squares: $x^2 - 25 = (x - 5)(x + 5)$.
3. Cancel the common factor $(x - 5)$ from the top and bottom: $\lim_{x \to 5} (x + 5)$.
4. Use direct substitution: $5 + 5 = 10$.

Answer: 10

1. Recognize that the function is a polynomial, so we can use direct substitution.
2. Substitute $x = 2$ into the expression: $4(2)^3 - 2(2) + 7$.
3. Calculate the powers: $4(8) - 4 + 7$.
4. Perform the multiplication and addition: $32 - 4 + 7 = 35$.

Answer: 35

1. Substitute $x = 1$ to check: $\frac{1 + 2 - 3}{1 - 1} = \frac{0}{0}$.
2. Factor the numerator: $x^2 + 2x - 3 = (x - 1)(x + 3)$.
3. Factor the denominator: $x^2 - 1 = (x - 1)(x + 1)$.
4. Cancel the common factor $(x - 1)$: $\lim_{x \to 1} \frac{x + 3}{x + 1}$.
5. Substitute $x = 1$: $\frac{1 + 3}{1 + 1} = \frac{4}{2} = 2$.

Answer: 2

1. Start with the known bound for sine: $-1 \le \sin\left(\frac{5}{x}\right) \le 1$.
2. Multiply the entire inequality by $x^2$. Since $x^2 \ge 0$ for all $x$, the inequality signs remain the same: $-x^2 \le x^2 \sin\left(\frac{5}{x}\right) \le x^2$.
3. Evaluate the limit of the lower bound: $\lim_{x \to 0} -x^2 = 0$.
4. Evaluate the limit of the upper bound: $\lim_{x \to 0} x^2 = 0$.
5. Since both bounds approach $0$, by the Squeeze Theorem, the middle limit must also be $0$.

Answer: 0

1. Identify the lower bound function $g(x) = 2x - 1$ and the upper bound function $h(x) = x^2 + 1$.
2. Calculate the limit of the lower bound as $x$ approaches $1$: $\lim_{x \to 1} (2x - 1) = 2(1) - 1 = 1$.
3. Calculate the limit of the upper bound as $x$ approaches $1$: $\lim_{x \to 1} (x^2 + 1) = 1^2 + 1 = 2$.
4. Observe that the limits are $1$ and $2$. Since $1 \neq 2$, the Squeeze Theorem cannot be used to find a specific value for the limit.

Answer: The limit cannot be determined using the Squeeze Theorem because the bounds do not approach the same value.

1. We want the expression inside the sine to match the denominator. We have $3x$ inside the sine, but only $x$ in the denominator.
2. Multiply the expression by $\frac{3}{3}$ to manipulate the denominator: $\lim_{x \to 0} 3 \cdot \frac{\sin(3x)}{3x}$.
3. Use the constant multiple law to move the $3$ outside the limit: $3 \cdot \lim_{x \to 0} \frac{\sin(3x)}{3x}$.
4. As $x$ approaches $0$, the term $3x$ also approaches $0$. Therefore, $\lim_{x \to 0} \frac{\sin(3x)}{3x} = 1$ based on the fundamental limit proven today.
5. The final result is $3 \cdot 1 = 3$.

Answer: 3

1. Identify the highest power of $x$ in the denominator, which is $x^2$.
2. Divide every term in the numerator and denominator by $x^2$: $\frac{6 - 3/x + 1/x^2}{2 + 5/x^2}$.
3. Apply the limit as $x \to \infty$, noting that $3/x$, $1/x^2$, and $5/x^2$ all approach $0$.
4. The expression simplifies to $6/2 = 3$.

Answer: 3

1. Compare the degrees of the numerator and denominator. The numerator degree is $1$ and the denominator degree is $2$.
2. Since the degree of the denominator is higher, the limit as $x \to \infty$ is $0$.
3. Similarly, the limit as $x \to {-\infty}$ is $0$.
4. The horizontal asymptote is the line $y = 0$.

Answer: y = 0

1. Divide all terms by the highest power in the denominator, $x^3$.
2. The expression becomes $\frac{2 - 4/x^2}{1 + 7/x + 1/x^3}$.
3. As $x \to \infty$, the terms $4/x^2$, $7/x$, and $1/x^3$ all approach $0$.
4. The limit is $2/1 = 2$.

Answer: 2

1. Set the denominator to zero: $x + 5 = 0$, which gives $x = -5$.
2. Since the numerator is a non-zero constant, $x = -5$ is a vertical asymptote.
3. Test the right-hand limit: for $x = -4.9$, $f(-4.9) = \frac{4}{0.1} = 40$ (positive), so $\lim_{x \to -5^+} f(x) = \infty$.
4. Test the left-hand limit: for $x = -5.1$, $f(-5.1) = \frac{4}{-0.1} = -40$ (negative), so $\lim_{x \to -5^-} f(x) = -\infty$.

Answer: Vertical asymptote at $x = -5$; $\lim_{x \to -5^+} f(x) = \infty$ and $\lim_{x \to -5^-} f(x) = -\infty$.

1. The denominator is zero when $(x-2)^2 = 0$, which means $x = 2$.
2. The numerator is $1$, so $x = 2$ is a vertical asymptote.
3. Test the right-hand limit: for $x = 2.1$, $(2.1-2)^2 = 0.01$, and $1/0.01 = 100$ (positive), so $\lim_{x \to 2^+} f(x) = \infty$.
4. Test the left-hand limit: for $x = 1.9$, $(1.9-2)^2 = (-0.1)^2 = 0.01$, and $1/0.01 = 100$ (positive), so $\lim_{x \to 2^-} f(x) = \infty$.

Answer: Vertical asymptote at $x = 2$; $\lim_{x \to 2^+} f(x) = \infty$ and $\lim_{x \to 2^-} f(x) = \infty$.

1. Factor the denominator: $x^2 - 5x + 6 = (x-2)(x-3)$.
2. The function is $f(x) = \frac{x-3}{(x-2)(x-3)}$.
3. The value $x = 3$ makes both numerator and denominator zero, so it is a hole.
4. The value $x = 2$ makes only the denominator zero, so it is a vertical asymptote.
5. Simplify the function to $f(x) = \frac{1}{x-2}$ for $x \neq 3$ to confirm the behavior at $x=2$.

Answer: The only vertical asymptote is $x = 2$.

1. Find the function value: $f(2) = 2^2 = 4$.
2. Find the left-hand limit: $\lim_{x \to 2^-} x^2 = 4$.
3. Find the right-hand limit: $\lim_{x \to 2^+} 2x = 2(2) = 4$.
4. Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 4$, the function is continuous.

Answer: Continuous

1. Check the function value: $f(3) = \frac{0}{0}$, which is undefined.
2. Compute the limit: $\lim_{x \to 3} \frac{x-3}{(x-3)(x+3)} = \lim_{x \to 3} \frac{1}{x+3} = \frac{1}{6}$.
3. Since the limit exists as a finite number but the function is undefined, it is a removable discontinuity.

Answer: Removable Discontinuity

1. Check the function value: $f(5) = \frac{1}{0}$, which is undefined.
2. Check the right-hand limit: $\lim_{x \to 5^+} \frac{1}{x-5} = \infty$.
3. Check the left-hand limit: $\lim_{x \to 5^-} \frac{1}{x-5} = -\infty$.
4. Since the limits are infinite, this is an infinite discontinuity.

Answer: Infinite Discontinuity

1. Step 1: Note that $f(x)$ is a polynomial, so it is continuous on the closed interval $[1, 2]$.
2. Step 2: Evaluate the function at the left endpoint: $f(1) = 1^3 + 2(1) - 5 = 1 + 2 - 5 = -2$.
3. Step 3: Evaluate the function at the right endpoint: $f(2) = 2^3 + 2(2) - 5 = 8 + 4 - 5 = 7$.
4. Step 4: Since $f(1) = -2$ and $f(2) = 7$, and $0$ is between $-2$ and $7$, the Intermediate Value Theorem guarantees there is at least one $c$ in $(1, 2)$ such that $f(c) = 0$.

Answer: A root exists because $f(1) = -2$ and $f(2) = 7$ and the function is continuous.

1. Step 1: Define a new function $g(x) = \tan(x) - 2x$. We want to find where $g(c) = 0$.
2. Step 2: Check continuity. $\tan(x)$ is continuous on $[1, 1.5]$ because the first vertical asymptote is at $\pi/2 \approx 1.57$, which is outside this interval.
3. Step 3: Evaluate $g(1) = \tan(1) - 2(1) \approx 1.557 - 2 = -0.443$.
4. Step 4: Evaluate $g(1.5) = \tan(1.5) - 2(1.5) \approx 14.101 - 3 = 11.101$.
5. Step 5: Since $g(1) < 0$ and $g(1.5) > 0$, the IVT guarantees a root in $(1, 1.5)$.

Answer: A root exists in $(1, 1.5)$ because $g(1) \approx -0.443$ and $g(1.5) \approx 11.101$ and the function is continuous.

1. Step 1: $f(x)$ is continuous on $[0, 2]$ as it is the difference of an exponential and a linear function.
2. Step 2: Evaluate at $x=0$: $f(0) = e^0 - 3(0) = 1 - 0 = 1$ (positive).
3. Step 3: Evaluate at $x=1$: $f(1) = e^1 - 3(1) \approx 2.718 - 3 = -0.282$ (negative).
4. Step 4: Evaluate at $x=2$: $f(2) = e^2 - 3(2) \approx 7.389 - 6 = 1.389$ (positive).
5. Step 5: There is a sign change between $x=0$ and $x=1$, so there is a root in $(0, 1)$.
6. Step 6: There is another sign change between $x=1$ and $x=2$, so there is a second root in $(1, 2)$.

Answer: Roots exist in $(0, 1)$ and $(1, 2)$ because the function values change signs twice.

1. Start with the output inequality: $|(4x - 5) - 7| < \epsilon$.
2. Simplify the expression: $|4x - 12| < \epsilon$.
3. Factor out the constant: $4|x - 3| < \epsilon$.
4. Isolate the input distance: $|x - 3| < \frac{\epsilon}{4}$.
5. Choose $\delta = \frac{\epsilon}{4}$. If $0 < |x - 3| < \delta$, then $4|x - 3| < \epsilon$, which implies $|(4x - 5) - 7| < \epsilon$.

Answer: \delta = \frac{\epsilon}{4}

1. Set up the inequality: $|(2x + 8) - 6| < \epsilon$.
2. Simplify: $|2x + 2| < \epsilon$.
3. Factor: $2|x + 1| < \epsilon$.
4. Solve for the distance from $x$ to $-1$: $|x - (-1)| < \frac{\epsilon}{2}$.
5. Choose $\delta = \frac{\epsilon}{2}$.

Answer: \delta = \frac{\epsilon}{2}

1. Start with $|(-5x + 2) - L| < \epsilon$.
2. Since $L = -5a + 2$, substitute this in: $|-5x + 2 - (-5a + 2)| < \epsilon$.
3. Simplify: $|-5x + 5a| < \epsilon$.
4. Factor out $-5$: $|-5(x - a)| < \epsilon$.
5. Use absolute value properties: $5|x - a| < \epsilon$.
6. Divide by 5: $|x - a| < \frac{\epsilon}{5}$.

Answer: \delta = \frac{\epsilon}{5}

1. Start with $|(3x + 2) - 14| < \epsilon$.
2. Simplify to $|3x - 12| < \epsilon$, which is $3|x - 4| < \epsilon$.
3. This suggests $|x - 4| < \epsilon/3$.
4. Choose $\delta = \epsilon/3$.
5. Verify: if $|x - 4| < \delta$, then $3|x - 4| < 3(\epsilon/3) = \epsilon$, so $|(3x + 2) - 14| < \epsilon$.

Answer: \delta = \frac{\epsilon}{3}

1. Start with $|x^2 - 9| = |x - 3||x + 3| < \epsilon$.
2. Assume $|x - 3| < 1$, so $2 < x < 4$.
3. Then $|x + 3| < 4 + 3 = 7$.
4. We need $|x - 3| \cdot 7 < \epsilon$, so $|x - 3| < \epsilon/7$.
5. Choose $\delta = \min(1, \epsilon/7)$.

Answer: \delta = \min(1, \frac{\epsilon}{7})

1. Start with $|(x^2 + 2x) - 3| = |(x - 1)(x + 3)| < \epsilon$.
2. Assume $|x - 1| < 1$, so $0 < x < 2$.
3. Then $|x + 3| < 2 + 3 = 5$.
4. We need $|x - 1| \cdot 5 < \epsilon$, so $|x - 1| < \epsilon/5$.
5. Choose $\delta = \min(1, \epsilon/5)$.

Answer: \delta = \min(1, \frac{\epsilon}{5})

1. Let $m = \frac{n}{3}$, which implies $n = 3m$.
2. As $n \to \infty$, $m \to \infty$.
3. Rewrite the limit as $\lim_{m \to \infty} (1 + \frac{1}{m})^{3m}$.
4. Apply the power rule for limits: $[\lim_{m \to \infty} (1 + \frac{1}{m})^m]^3$.
5. Substitute the definition of $e$: $e^3$.

Answer: e^3

1. Let $m = 2n$.
2. As $n \to \infty$, $m \to \infty$.
3. Substitute $m$ into the expression: $\lim_{m \to \infty} (1 + \frac{1}{m})^m$.
4. This is the exact definition of $e$.

Answer: e

1. Rewrite the expression as $(1 + \frac{-2}{n})^n$.
2. Let $m = \frac{n}{-2}$, so $n = -2m$.
3. As $n \to \infty$, $m \to -\infty$.
4. The limit becomes $\lim_{m \to -\infty} (1 + \frac{1}{m})^{-2m}$.
5. This is $[\lim_{m \to -\infty} (1 + \frac{1}{m})^m]^{-2}$.
6. Substitute $e$: $e^{-2}$ or $\frac{1}{e^2}$.

Answer: e^{-2}

1. Identify that the argument of sine is $7x$ but the denominator is $2x$.
2. Multiply the numerator and denominator by $7$ to match the argument: $\frac{7}{2} \cdot \frac{\sin(7x)}{7x}$.
3. Apply the limit $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$ where $u = 7x$.
4. The result is $\frac{7}{2} \cdot 1 = 3.5$.

Answer: 3.5

1. Identify that the argument of cosine is $4x$.
2. To use the formula $\lim_{u \to 0} \frac{1 - \cos(u)}{u} = 0$, we need $4x$ in the denominator.
3. Multiply the top and bottom by $4$: $4 \cdot \frac{1 - \cos(4x)}{4x}$.
4. The limit of the fraction part is $0$, so $4 \cdot 0 = 0$.

Answer: 0

1. Split the limit into two separate fractions: $\lim_{x \to 0} \frac{\sin(3x)}{x} + \lim_{x \to 0} \frac{1 - \cos(x)}{x}$.
2. For the first part, multiply by $3/3$ to get $3 \cdot \lim_{x \to 0} \frac{\sin(3x)}{3x} = 3 \cdot 1 = 3$.
3. For the second part, use the known limit $\lim_{x \to 0} \frac{1 - \cos(x)}{x} = 0$.
4. Add the results: $3 + 0 = 3$.

Answer: 3

1. Plugging in $x = -2$ gives $\frac{(-2)^2 + 5(-2) + 6}{-2 + 2} = \frac{4 - 10 + 6}{0} = \frac{0}{0}$, which is an indeterminate form.
2. Factor the numerator: $x^2 + 5x + 6 = (x + 2)(x + 3)$.
3. Rewrite the limit: $\lim_{x \to -2} \frac{(x + 2)(x + 3)}{x + 2}$.
4. Cancel the common factor $(x + 2)$ from the top and bottom: $\lim_{x \to -2} (x + 3)$.
5. Substitute $x = -2$ into the simplified expression: $-2 + 3 = 1$.

Answer: 1

1. Identify the highest power of $x$ in the denominator, which is $x^2$.
2. Divide every term by $x^2$: $\lim_{x \to \infty} \frac{3 - \frac{2}{x} + \frac{1}{x^2}}{5 + \frac{7}{x^2}}$.
3. Apply the limit laws: $\frac{2}{x}$, $\frac{1}{x^2}$, and $\frac{7}{x^2}$ all approach $0$ as $x$ approaches infinity.
4. The expression becomes $\frac{3 - 0 + 0}{5 + 0} = \frac{3}{5}$.

Answer: \frac{3}{5}

1. Check the limit from the right ($x > 0$): $\lim_{x \to 0^+} \frac{x}{x(x + 1)} = \lim_{x \to 0^+} \frac{1}{x + 1} = 1$.
2. Check the limit from the left ($x < 0$): $\lim_{x \to 0^-} \frac{-x}{x(x + 1)} = \lim_{x \to 0^-} \frac{-1}{x + 1} = -1$.
3. Since the left-hand limit ($-1$) and right-hand limit ($1$) are not equal, the two-sided limit does not exist.

Answer: \text{Does Not Exist}

1. Substitute $x = 4$ to get $\frac{16-16}{4-4} = \frac{0}{0}$, an indeterminate form.
2. Factor the numerator as a difference of squares: $x^2 - 16 = (x - 4)(x + 4)$.
3. Cancel the common factor $(x - 4)$ from the top and bottom.
4. Evaluate the remaining limit: $\lim_{x \to 4} (x + 4) = 4 + 4 = 8$.

Answer: 8

1. Identify the form $\frac{\sin(ax)}{bx}$.
2. Rewrite the expression to isolate the $\frac{\sin(7x)}{7x}$ form: $\frac{7}{2} \cdot \frac{\sin(7x)}{7x}$.
3. Apply the special limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$ where $u = 7x$.
4. The result is $\frac{7}{2} \cdot 1 = 3.5$.

Answer: 7/2

1. Identify the highest power of $x$ in the denominator, which is $x^2$.
2. Divide every term by $x^2$: $\frac{3 - 5/x^2}{2 + 1/x + 1/x^2}$.
3. Take the limit as $x \to \infty$, noting that $5/x^2$, $1/x$, and $1/x^2$ all approach $0$.
4. The limit is $\frac{3 - 0}{2 + 0 + 0} = \frac{3}{2}$.

Answer: y = 3/2

1. Set up the limit: $\lim_{h \to 0} \frac{(1+h)^2 + 2(1+h) - (1^2 + 2(1))}{h}$
2. Expand the numerator: $\lim_{h \to 0} \frac{1 + 2h + h^2 + 2 + 2h - 3}{h}$
3. Simplify the numerator: $\lim_{h \to 0} \frac{4h + h^2}{h}$
4. Factor and cancel $h$: $\lim_{h \to 0} (4 + h)$
5. Evaluate the limit as $h \to 0$: $4 + 0 = 4$

Answer: 4

1. Set up the limit: $\lim_{h \to 0} \frac{((2+h)^2 - 3(2+h)) - (2^2 - 3(2))}{h}$
2. Expand the terms: $\lim_{h \to 0} \frac{4 + 4h + h^2 - 6 - 3h - (4 - 6)}{h}$
3. Simplify the numerator: $\lim_{h \to 0} \frac{h^2 + h - 2 - (-2)}{h} = \lim_{h \to 0} \frac{h^2 + h}{h}$
4. Factor and cancel $h$: $\lim_{h \to 0} (h + 1)$
5. Evaluate the limit: $0 + 1 = 1$

Answer: 1

1. Set up the limit: $\lim_{h \to 0} \frac{3(-1+h)^2 - 5 - (3(-1)^2 - 5)}{h}$
2. Expand the squared term: $\lim_{h \to 0} \frac{3(1 - 2h + h^2) - 5 - (3 - 5)}{h}$
3. Simplify the numerator: $\lim_{h \to 0} \frac{3 - 6h + 3h^2 - 5 - (-2)}{h}$
4. Combine constants: $\lim_{h \to 0} \frac{-6h + 3h^2}{h}$
5. Factor and cancel $h$: $\lim_{h \to 0} (-6 + 3h)$
6. Evaluate the limit: $-6$

Answer: -6

1. Set up the limit: $v(1) = \lim_{h \to 0} \frac{s(1+h) - s(1)}{h}$
2. Calculate $s(1) = 3(1)^2 - 2(1) = 1$
3. Calculate $s(1+h) = 3(1+h)^2 - 2(1+h) = 3(1 + 2h + h^2) - 2 - 2h = 3 + 6h + 3h^2 - 2 - 2h = 1 + 4h + 3h^2$
4. Substitute into the limit: $\lim_{h \to 0} \frac{(1 + 4h + 3h^2) - 1}{h} = \lim_{h \to 0} \frac{4h + 3h^2}{h}$
5. Simplify: $\lim_{h \to 0} (4 + 3h) = 4$

Answer: 4 \text{ m/s}

1. Set up the limit for the derivative at $x=10$: $C'(10) = \lim_{h \to 0} \frac{C(10+h) - C(10)}{h}$
2. Calculate $C(10) = 0.5(100) + 10(10) + 100 = 50 + 100 + 100 = 250$
3. Calculate $C(10+h) = 0.5(10+h)^2 + 10(10+h) + 100 = 0.5(100 + 20h + h^2) + 100 + 10h + 100 = 50 + 10h + 0.5h^2 + 200 + 10h = 250 + 20h + 0.5h^2$
4. Substitute into the limit: $\lim_{h \to 0} \frac{(250 + 20h + 0.5h^2) - 250}{h} = \lim_{h \to 0} \frac{20h + 0.5h^2}{h}$
5. Simplify: $\lim_{h \to 0} (20 + 0.5h) = 20$

Answer: 20 \text{ dollars per unit}

1. Find the general derivative $s'(t)$: $\lim_{h \to 0} \frac{2(t+h)^2 - 2t^2}{h}$
2. Expand $2(t+h)^2 = 2(t^2 + 2th + h^2) = 2t^2 + 4th + 2h^2$
3. Simplify the limit: $\lim_{h \to 0} \frac{4th + 2h^2}{h} = \lim_{h \to 0} (4t + 2h) = 4t$
4. Set the velocity equal to 12: $4t = 12$
5. Solve for $t$: $t = 3$

Answer: t = 3 \text{ seconds}

1. Check continuity: $f(0) = 0^2 + 1 = 1$. The limit from the right is $0 + 1 = 1$. It is continuous.
2. Find the left-hand derivative: $\lim_{h \to 0^-} \frac{(0+h)^2+1 - (0^2+1)}{h} = \lim_{h \to 0^-} \frac{h^2}{h} = 0$.
3. Find the right-hand derivative: $\lim_{h \to 0^+} \frac{(0+h)+1 - (0+1)}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$.
4. Compare the slopes: $0 \neq 1$.
5. Since the one-sided derivatives are different, the function is not differentiable at $x = 0$.

Answer: Not differentiable at $x = 0$ (it has a corner).

1. Check continuity: $f(3) = |2(3) - 6| = 0$. The function is continuous.
2. For $x > 3$, $f(x) = 2x - 6$. The difference quotient is $\frac{2(3+h)-6 - 0}{h} = \frac{2h}{h} = 2$.
3. For $x < 3$, $f(x) = -(2x - 6) = -2x + 6$. The difference quotient is $\frac{-2(3+h)+6 - 0}{h} = \frac{-2h}{h} = -2$.
4. The left-hand slope $-2$ does not equal the right-hand slope $2$.

Answer: Not differentiable at $x = 3$.

1. Check continuity: $f(0) = 0^{1/5} = 0$. It is continuous.
2. Set up the difference quotient: $\lim_{h \to 0} \frac{h^{1/5} - 0}{h}$.
3. Simplify the expression: $\frac{h^{1/5}}{h} = h^{1/5 - 1} = h^{-4/5} = \frac{1}{h^{4/5}}$.
4. Evaluate the limit: as $h \to 0$, $1/h^{4/5}$ approaches infinity.
5. Since the slope becomes infinite, there is a vertical tangent.

Answer: It has a vertical tangent at $x = 0$.

1. Identify the exponent $n = 7$.
2. Apply the power rule: bring the $7$ to the front and subtract $1$ from the exponent.
3. The result is $7x^{7-1}$.

Answer: f'(x) = 7x^6

1. Rewrite the function using a negative exponent: $f(x) = x^{-4}$.
2. Apply the power rule: bring $-4$ to the front and subtract $1$ from the exponent.
3. Compute $-4 - 1 = -5$.
4. The derivative is $-4x^{-5}$, which can be rewritten as $-\frac{4}{x^5}$.

Answer: f'(x) = -\frac{4}{x^5}

1. Rewrite the radical as a fractional exponent: $f(x) = (x^2)^{1/5} = x^{2/5}$.
2. Apply the power rule: bring $2/5$ to the front and subtract $1$ from the exponent.
3. Compute $2/5 - 1 = 2/5 - 5/5 = -3/5$.
4. The derivative is $\frac{2}{5}x^{-3/5}$.

Answer: f'(x) = \frac{2}{5}x^{-3/5}

1. Apply the power rule to $7x^4$ to get $7(4x^3) = 28x^3$.
2. Apply the power rule to $-2x^3$ to get $-2(3x^2) = -6x^2$.
3. Apply the power rule to $5x$ to get $5(1) = 5$.
4. The derivative of the constant $-12$ is $0$.

Answer: f'(x) = 28x^3 - 6x^2 + 5

1. Differentiate $\frac{1}{4}x^4$ as $\frac{1}{4}(4x^3) = x^3$.
2. Differentiate $\frac{2}{3}x^3$ as $\frac{2}{3}(3x^2) = 2x^2$.
3. The derivative of $-8$ is $0$.

Answer: g'(x) = x^3 + 2x^2

1. First, expand the expression: $h(x) = 2x^2 - 10x + x - 5 = 2x^2 - 9x - 5$.
2. Differentiate $2x^2$ to get $4x$.
3. Differentiate $-9x$ to get $-9$.
4. The derivative of $-5$ is $0$.

Answer: h'(x) = 4x - 9

1. Identify $u = x^3$ and $v = \ln x$.
2. Compute derivatives: $u' = 3x^2$ and $v' = \frac{1}{x}$.
3. Apply product rule: $f'(x) = (3x^2)(\ln x) + (x^3)(\frac{1}{x})$.
4. Simplify the second term: $x^3 \cdot \frac{1}{x} = x^2$.
5. Combine: $f'(x) = 3x^2 \ln x + x^2$.

Answer: f'(x) = x^2(3 \ln x + 1)

1. Identify $u = x^2$ and $v = \sin x$.
2. Compute derivatives: $u' = 2x$ and $v' = \cos x$.
3. Apply product rule: $g'(x) = (2x)(\sin x) + (x^2)(\cos x)$.
4. Factor out $x$: $g'(x) = x(2 \sin x + x \cos x)$.

Answer: g'(x) = 2x \sin x + x^2 \cos x

1. Identify $u = x^2 + 1$ and $v = e^x$.
2. Compute derivatives: $u' = 2x$ and $v' = e^x$.
3. Apply product rule: $h'(x) = (2x)(e^x) + (x^2 + 1)(e^x)$.
4. Factor out $e^x$: $h'(x) = e^x(2x + x^2 + 1)$.
5. Recognize the quadratic as a perfect square: $x^2 + 2x + 1 = (x+1)^2$.

Answer: h'(x) = (x+1)^2 e^x

1. Identify $f(x) = x+2$ and $g(x) = x-3$.
2. Compute derivatives: $f'(x) = 1$ and $g'(x) = 1$.
3. Apply quotient rule: $\frac{(1)(x-3) - (x+2)(1)}{(x-3)^2}$.
4. Simplify numerator: $x - 3 - x - 2 = -5$.

Answer: f'(x) = \frac{-5}{(x-3)^2}

1. Identify high as $x^2$ (derivative $2x$) and low as $2x+1$ (derivative $2$).
2. Apply rule: $\frac{(2x)(2x+1) - (x^2)(2)}{(2x+1)^2}$.
3. Expand numerator: $4x^2 + 2x - 2x^2$.
4. Simplify: $2x^2 + 2x$.

Answer: y' = \frac{2x^2 + 2x}{(2x+1)^2}

1. Rewriting as $g(x) = e^x \cdot x^{-2}$ allows the product rule.
2. Using product rule: $g'(x) = (e^x)(x^{-2}) + (e^x)(-2x^{-3})$.
3. Factor out $e^x$ and $x^{-3}$: $e^x x^{-3}(x - 2)$.
4. Convert back to fraction: $\frac{e^x(x-2)}{x^3}$.

Answer: g'(x) = \frac{e^x(x-2)}{x^3}

1. Apply the linearity of the derivative to separate the terms: $f'(x) = 3\frac{d}{dx}\sin(x) - 4\frac{d}{dx}\cos(x)$.
2. Use the known derivatives: $\frac{d}{dx}\sin(x) = \cos(x)$ and $\frac{d}{dx}\cos(x) = -\sin(x)$.
3. Substitute these into the expression: $f'(x) = 3\cos(x) - 4(-\sin(x))$.
4. Simplify the signs: $f'(x) = 3\cos(x) + 4\sin(x)$.

Answer: f'(x) = 3\cos(x) + 4\sin(x)

1. Identify this as a product of $u(x) = x^2$ and $v(x) = \cos(x)$.
2. Apply the product rule: $g'(x) = u'(x)v(x) + u(x)v'(x)$.
3. Compute the individual derivatives: $u'(x) = 2x$ and $v'(x) = -\sin(x)$.
4. Substitute back: $g'(x) = (2x)\cos(x) + (x^2)(-\sin(x))$.
5. Simplify the expression: $g'(x) = 2x\cos(x) - x^2\sin(x)$.

Answer: g'(x) = 2x\cos(x) - x^2\sin(x)

1. Identify this as a quotient of $u(x) = \cos(x)$ and $v(x) = x+1$.
2. Apply the quotient rule: $h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.
3. Compute the individual derivatives: $u'(x) = -\sin(x)$ and $v'(x) = 1$.
4. Substitute into the formula: $h'(x) = \frac{(-\sin(x))(x+1) - (\cos(x))(1)}{(x+1)^2}$.
5. Simplify the numerator: $h'(x) = \frac{-(x+1)\sin(x) - \cos(x)}{(x+1)^2}$.

Answer: h'(x) = \frac{-(x+1)\sin(x) - \cos(x)}{(x+1)^2}

1. Identify the constant multiple as $5$.
2. Recall that the derivative of $\sec(x)$ is $\sec(x)\tan(x)$.
3. Apply the linearity rule: $f'(x) = 5 \cdot \frac{d}{dx}\sec(x)$.

Answer: f'(x) = 5\sec(x)\tan(x)

1. Differentiate $\tan(x)$ to get $\sec^2(x)$.
2. Differentiate $\cos(x)$ to get $-\sin(x)$.
3. Combine the results using the sum rule.

Answer: g'(x) = \sec^2(x) - \sin(x)

1. Use the product rule with $u = x^2$ and $v = \cot(x)$.
2. Compute $u' = 2x$ and $v' = -\csc^2(x)$.
3. Apply the formula $h'(x) = u'v + uv'$.
4. Substitute the values: $h'(x) = (2x)\cot(x) + (x^2)(-\csc^2(x))$.

Answer: h'(x) = 2x\cot(x) - x^2\csc^2(x)

1. Identify the outer function $f(u) = \ln(u)$ and the inner function $u = x^2 + 3$.
2. Find the derivative of the outer function: $f'(u) = \frac{1}{u}$.
3. Find the derivative of the inner function: $u' = 2x$.
4. Apply the chain rule: $\frac{dy}{dx} = \frac{1}{x^2 + 3} \cdot 2x$.

Answer: \frac{2x}{x^2 + 3}

1. Identify the outer function as cosine and the inner function as $4x^3$.
2. The derivative of the outer function is $-\sin(u)$, evaluated at $u = 4x^3$.
3. The derivative of the inner function $4x^3$ is $12x^2$.
4. Multiply them together: $-\sin(4x^3) \cdot 12x^2$.

Answer: -12x^2\sin(4x^3)

1. Identify the outer function as $u^7$ and the inner function as $u = 5x^2 - 2$.
2. Differentiate the outer layer: $7u^6$, which is $7(5x^2 - 2)^6$.
3. Differentiate the inner layer: $\frac{d}{dx}(5x^2 - 2) = 10x$.
4. Multiply the results: $7(5x^2 - 2)^6 \cdot 10x$.

Answer: 70x(5x^2 - 2)^6

1. Identify this as a product of $u = x$ and $v = \ln(x^2 + 1)$.
2. The derivative of $u$ is $u' = 1$.
3. The derivative of $v$ requires the chain rule: $v' = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}$.
4. Apply the product rule: $y' = u'v + uv' = 1 \cdot \ln(x^2 + 1) + x \cdot \frac{2x}{x^2 + 1}$.
5. Simplify the expression to $\ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}$.

Answer: \frac{dy}{dx} = \ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}

1. Rewrite the function as $y = (\sin(3x))^4$.
2. The outermost layer is the power of 4. Its derivative is $4(\sin(3x))^3$.
3. The next layer is the sine function. Its derivative is $\cos(3x)$.
4. The innermost layer is $3x$. Its derivative is $3$.
5. Multiply all layers together: $y' = 4\sin^3(3x) \cdot \cos(3x) \cdot 3$.
6. Simplify to $12\sin^3(3x)\cos(3x)$.

Answer: \frac{dy}{dx} = 12\sin^3(3x)\cos(3x)

1. Identify this as a quotient with $u = e^{2x}$ and $v = \cos(x)$.
2. Find $u'$ using the chain rule: $u' = e^{2x} \cdot 2 = 2e^{2x}$.
3. Find $v'$: $v' = -\sin(x)$.
4. Apply the quotient rule: $y' = \frac{u'v - uv'}{v^2} = \frac{(2e^{2x})(\cos(x)) - (e^{2x})(-\sin(x))}{(\cos(x))^2}$.
5. Factor out $e^{2x}$ in the numerator: $y' = \frac{e^{2x}(2\cos(x) + \sin(x))}{\cos^2(x)}$.

Answer: \frac{dy}{dx} = \frac{e^{2x}(2\cos(x) + \sin(x))}{\cos^2(x)}

1. Identify the base $a = 7$.
2. Apply the formula $\frac{d}{dx} a^x = a^x \ln a$.
3. Substitute $7$ for $a$ to get $7^x \ln 7$.

Answer: f'(x) = 7^x \ln 7

1. Identify the outer function as $e^u$ and the inner function as $u = 5x^3$.
2. The derivative of the outer function is $e^{5x^3}$.
3. The derivative of the inner function is $15x^2$.
4. Multiply them together using the chain rule: $15x^2 e^{5x^3}$.

Answer: f'(x) = 15x^2 e^{5x^3}

1. Use the product rule with $u = x$ and $v = e^{2x}$.
2. Compute $u' = 1$ and $v' = 2e^{2x}$ using the chain rule.
3. Apply the product rule: $f'(x) = u'v + uv' = (1)(e^{2x}) + (x)(2e^{2x})$.
4. Factor out $e^{2x}$ to simplify: $e^{2x}(1 + 2x)$.

Answer: f'(x) = e^{2x}(1 + 2x)

1. Identify the inner function $u = x^3 + 7x$.
2. Compute the derivative of the inner function: $u' = 3x^2 + 7$.
3. Apply the formula $\frac{d}{dx} \ln(u) = \frac{u'}{u}$.
4. Substitute the expressions: $f'(x) = \frac{3x^2 + 7}{x^3 + 7x}$.

Answer: $f'(x) = \frac{3x^2 + 7}{x^3 + 7x}$

1. Take the natural log of both sides: $\ln y = \ln(x^5 \sin x)$.
2. Use log laws to expand: $\ln y = 5 \ln x + \ln(\sin x)$.
3. Differentiate implicitly: $\frac{1}{y} \frac{dy}{dx} = \frac{5}{x} + \frac{\cos x}{\sin x}$.
4. Simplify the trig term: $\frac{1}{y} \frac{dy}{dx} = \frac{5}{x} + \cot x$.
5. Multiply by $y$: $\frac{dy}{dx} = (x^5 \sin x) (\frac{5}{x} + \cot x)$.

Answer: $\frac{dy}{dx} = (x^5 \sin x) (\frac{5}{x} + \cot x)$

1. Set $y = x^{\cos x}$ and take the natural log: $\ln y = \ln(x^{\cos x})$.
2. Bring the exponent down: $\ln y = (\cos x) \ln x$.
3. Differentiate using the product rule: $\frac{1}{y} \frac{dy}{dx} = (-\sin x)\ln x + (\cos x)\frac{1}{x}$.
4. Isolate $\frac{dy}{dx}$ by multiplying by $y$: $\frac{dy}{dx} = x^{\cos x} (\frac{\cos x}{x} - \sin x \ln x)$.

Answer: $f'(x) = x^{\cos x} (\frac{\cos x}{x} - \sin x \ln x)$

1. Identify the outer function as $\arcsin(u)$ and the inner function as $u = \sqrt{x}$.
2. The derivative of $\arcsin(u)$ is $\frac{1}{\sqrt{1-u^2}}$.
3. The derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$.
4. Apply the chain rule: $f'(x) = \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{1}{2\sqrt{x}}$.
5. Simplify the expression: $f'(x) = \frac{1}{2\sqrt{x}\sqrt{1-x}}$.

Answer: f'(x) = \frac{1}{2\sqrt{x-x^2}}

1. Identify the outer function as $\arctan(u)$ and the inner function as $u = e^{2x}$.
2. The derivative of $\arctan(u)$ is $\frac{1}{1+u^2}$.
3. The derivative of $e^{2x}$ is $2e^{2x}$ using the chain rule.
4. Apply the chain rule: $g'(x) = \frac{1}{1+(e^{2x})^2} \cdot 2e^{2x}$.
5. Simplify the exponent: $(e^{2x})^2 = e^{4x}$.

Answer: g'(x) = \frac{2e^{2x}}{1+e^{4x}}

1. Identify the outer function as $\ln(u)$ and the inner function as $u = \arctan(x)$.
2. The derivative of $\ln(u)$ is $\frac{1}{u}$.
3. The derivative of $\arctan(x)$ is $\frac{1}{1+x^2}$.
4. Apply the chain rule: $h'(x) = \frac{1}{\arctan(x)} \cdot \frac{1}{1+x^2}$.

Answer: h'(x) = \frac{1}{(1+x^2)\arctan(x)}

1. Differentiate both sides with respect to $x$: $2x + 6y \frac{dy}{dx} = 0$.
2. Subtract $2x$ from both sides: $6y \frac{dy}{dx} = -2x$.
3. Divide by $6y$: $\frac{dy}{dx} = \frac{-2x}{6y}$.
4. Simplify the fraction: $\frac{dy}{dx} = -\frac{x}{3y}$.

Answer: \frac{dy}{dx} = -\frac{x}{3y}

1. Apply the product rule to $x^2 y$: $(2xy + x^2 \frac{dy}{dx})$.
2. Apply the product rule to $y^2 x$: $(2yx \frac{dy}{dx} + y^2)$.
3. Combine: $2xy + x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + y^2 = 0$.
4. Group $\frac{dy}{dx}$ terms: $\frac{dy}{dx}(x^2 + 2xy) = -2xy - y^2$.
5. Isolate $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2xy}$.

Answer: \frac{dy}{dx} = -\frac{2xy + y^2}{x^2 + 2xy}

1. Differentiate both sides: $e^y \frac{dy}{dx} + 1 = \frac{dy}{dx}$.
2. Move $\frac{dy}{dx}$ terms to one side: $e^y \frac{dy}{dx} - \frac{dy}{dx} = -1$.
3. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(e^y - 1) = -1$.
4. Isolate $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-1}{e^y - 1}$.
5. Alternatively, multiply top and bottom by $-1$: $\frac{dy}{dx} = \frac{1}{1 - e^y}$.

Answer: \frac{dy}{dx} = \frac{1}{1 - e^y}

1. First, find the first derivative using the power rule: $f'(x) = 9x^2 - 14x$.
2. Next, differentiate the first derivative: $f''(x) = 18x - 14$.

Answer: f''(x) = 18x - 14

1. The first derivative of $\cos(x)$ is $-\sin(x)$.
2. The derivative of $-\sin(x)$ is $-\cos(x)$.

Answer: g''(x) = -\cos(x)

1. The first derivative is $h'(x) = 5e^{5x}$ using the chain rule.
2. The second derivative is $h''(x) = 5 \cdot 5e^{5x} = 25e^{5x}$.
3. The third derivative is $h'''(x) = 25 \cdot 5e^{5x} = 125e^{5x}$.

Answer: h'''(x) = 125e^{5x}

1. Identify the outer rule as the product rule with $u = x^3$ and $v = \cos(x^2)$.
2. Compute $u' = 3x^2$.
3. Compute $v'$ using the chain rule: $v' = -\sin(x^2) \cdot 2x = -2x \sin(x^2)$.
4. Apply the product rule: $f'(x) = (3x^2)(\cos(x^2)) + (x^3)(-2x \sin(x^2))$.
5. Simplify the expression: $f'(x) = 3x^2 \cos(x^2) - 2x^4 \sin(x^2)$.

Answer: f'(x) = 3x^2 \cos(x^2) - 2x^4 \sin(x^2)

1. Identify the outer rule as the quotient rule with $u = \ln(x)$ and $v = x^2$.
2. Compute $u' = 1/x$ and $v' = 2x$.
3. Apply the quotient rule: $g'(x) = \frac{(1/x)(x^2) - (\ln x)(2x)}{(x^2)^2}$.
4. Simplify the numerator: $(1/x)(x^2) = x$, so we have $\frac{x - 2x \ln x}{x^4}$.
5. Factor out $x$ from the numerator and cancel with the denominator: $g'(x) = \frac{x(1 - 2 \ln x)}{x^4} = \frac{1 - 2 \ln x}{x^3}$.

Answer: g'(x) = \frac{1 - 2 \ln x}{x^3}

1. Identify the outer rule as the chain rule for the exponential function $e^u$ where $u = x \sin(x)$.
2. The derivative is $h'(x) = e^{x \sin(x)} \cdot \frac{d}{dx}(x \sin(x))$.
3. To find the derivative of $x \sin(x)$, use the product rule: $(1)(\sin x) + (x)(\cos x)$.
4. Combine the results: $h'(x) = e^{x \sin(x)} (\sin x + x \cos x)$.

Answer: h'(x) = e^{x \sin(x)} (\sin x + x \cos x)

1. The relation is $A = s^2$.
2. Differentiate with respect to $t$: $\frac{dA}{dt} = 2s \frac{ds}{dt}$.
3. Substitute $s = 8$ and $\frac{ds}{dt} = 5$.
4. $\frac{dA}{dt} = 2(8)(5) = 80$.

Answer: 80 \text{ cm}^2/\text{s}

1. The relation is $V = \frac{4}{3}\pi r^3$.
2. Differentiate with respect to $t$: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
3. Substitute $r = 3$ and $\frac{dr}{dt} = 2$.
4. $\frac{dV}{dt} = 4\pi(3^2)(2) = 4\pi(9)(2) = 72\pi$.

Answer: 72\pi \text{ cm}^3/\text{s}

1. The relation is $x^2 + y^2 = z^2$.
2. Differentiate with respect to $t$: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}$.
3. Find $z$ when $x=3, y=4$: $z = \sqrt{3^2 + 4^2} = 5$.
4. Substitute $\frac{dx}{dt} = 3$ and $\frac{dy}{dt} = -4$ (since it moves toward the origin).
5. $2(3)(3) + 2(4)(-4) = 2(5) \frac{dz}{dt} \implies 18 - 32 = 10 \frac{dz}{dt} \implies -14 = 10 \frac{dz}{dt}$.
6. $\frac{dz}{dt} = -1.4$.

Answer: -1.4 \text{ units/s}

1. The area of a circle is $A = \pi r^2$.
2. Differentiate with respect to time $t$: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
3. Substitute the given values $r = 5$ and $\frac{dr}{dt} = 2$.
4. $\frac{dA}{dt} = 2\pi(5)(2) = 20\pi$.

Answer: $20\pi\text{ cm}^2\text{/s}$

1. Equation: $x^2 + y^2 = 13^2$.
2. Differentiate: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$.
3. Find $x$ when $y=5$: $x^2 + 5^2 = 13^2 \implies x^2 = 144 \implies x = 12$.
4. Substitute $x=12, y=5, \frac{dy}{dt}=-1$: $2(12)\frac{dx}{dt} + 2(5)(-1) = 0$.
5. $24\frac{dx}{dt} - 10 = 0 \implies \frac{dx}{dt} = \frac{10}{24} = \frac{5}{12}$.

Answer: $\frac{5}{12}\text{ ft/s}$

1. Volume $V = \frac{1}{3}\pi r^2 h$. Ratio $\frac{r}{h} = \frac{4}{10} = \frac{2}{5}$, so $r = \frac{2}{5}h$.
2. Substitute $r$: $V = \frac{1}{3}\pi (\frac{2}{5}h)^2 h = \frac{4}{75}\pi h^3$.
3. Differentiate: $\frac{dV}{dt} = \frac{4}{25}\pi h^2 \frac{dh}{dt}$.
4. Substitute $\frac{dV}{dt} = 2$ and $h = 5$: $2 = \frac{4}{25}\pi (5)^2 \frac{dh}{dt}$.
5. $2 = 4\pi \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{1}{2\pi}$.

Answer: $\frac{1}{2\pi}\text{ m/min}$

1. Let $f(x) = x^{1/3}$. We choose $a = 27$ because it is a perfect cube near $28$.
2. Calculate $f(27) = 27^{1/3} = 3$.
3. Find the derivative: $f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3(\sqrt[3]{x})^2}$.
4. Evaluate the derivative at $a = 27$: $f'(27) = \frac{1}{3(3^2)} = \frac{1}{27}$.
5. Apply the formula: $L(28) = f(27) + f'(27)(28 - 27) = 3 + \frac{1}{27}(1)$.
6. Simplify to get $3 + 0.037 = 3.037$.

Answer: 3.037

1. The volume of a sphere is $V = \frac{4}{3}\pi r^3$.
2. The differential is $dV = V'(r) dr = 4\pi r^2 dr$.
3. Substitute the known values: $r = 5$ and $dr = 0.1$.
4. Calculate $dV = 4\pi(5^2)(0.1) = 4\pi(25)(0.1)$.
5. Simplify to $100\pi(0.1) = 10\pi$.

Answer: 10\pi \text{ cm}^3 \approx 31.4 \text{ cm}^3

1. The area of a square is $A = x^2$.
2. The differential is $dA = 2x dx$.
3. Substitute the values: $x = 10$ and $dx = 0.05$.
4. Calculate $dA = 2(10)(0.05)$.
5. Simplify to $20(0.05) = 1$.

Answer: 1 \text{ cm}^2

1. Check the form: as $x \to 0$, $x^2 \to 0$ and $\sin(x) \to 0$. This is a $0/0$ form.
2. Apply L'Hopital's rule: differentiate the numerator to get $2x$ and the denominator to get $\cos(x)$.
3. Evaluate the new limit: $\lim_{x \to 0} \frac{2x}{\cos(x)}$.
4. Substitute $x = 0$: $\frac{2(0)}{\cos(0)} = \frac{0}{1} = 0$.

Answer: 0

1. Check the form: $e^0 - 1 - 0 = 0$ and $0^2 = 0$. This is $0/0$.
2. Apply L'Hopital's rule: the derivative of $e^x - 1 - x$ is $e^x - 1$, and the derivative of $x^2$ is $2x$.
3. Check the new form: $e^0 - 1 = 0$ and $2(0) = 0$. This is still $0/0$.
4. Apply L'Hopital's rule again: the derivative of $e^x - 1$ is $e^x$, and the derivative of $2x$ is $2$.
5. Evaluate the limit: $\lim_{x \to 0} \frac{e^x}{2} = \frac{e^0}{2} = \frac{1}{2}$.

Answer: 1/2

1. Check the form: $\tan(0) - 0 = 0$ and $0^3 = 0$. This is $0/0$.
2. Apply L'Hopital's rule: the derivative of $\tan(x) - x$ is $\sec^2(x) - 1$, and the derivative of $x^3$ is $3x^2$.
3. Note that $\sec^2(x) - 1 = \tan^2(x)$. The limit is $\lim_{x \to 0} \frac{\tan^2(x)}{3x^2}$.
4. This is still $0/0$. Apply L'Hopital's rule again: the derivative of $\tan^2(x)$ is $2\tan(x)\sec^2(x)$, and the derivative of $3x^2$ is $6x$.
5. The limit is $\lim_{x \to 0} \frac{2\tan(x)\sec^2(x)}{6x} = \lim_{x \to 0} \frac{\tan(x)}{3x} \cdot \sec^2(x)$.
6. Apply L'Hopital's rule one last time to $\frac{\tan(x)}{3x}$: the derivative of $\tan(x)$ is $\sec^2(x)$ and the derivative of $3x$ is $3$.
7. The limit becomes $\frac{\sec^2(0)}{3} \cdot \sec^2(0) = \frac{1}{3} \cdot 1 = \frac{1}{3}$.

Answer: 1/3

1. Identify the form: as $x \to 0^+$, $x^2 \to 0$ and $\ln x \to -\infty$. This is a $0 \cdot \infty$ form.
2. Rewrite as a quotient: $\lim_{x \to 0^+} \frac{\ln x}{1/x^2}$. This is now $-\infty / \infty$.
3. Apply L'Hopital's rule: the derivative of $\ln x$ is $1/x$ and the derivative of $1/x^2$ is $-2/x^3$.
4. Simplify the expression: $\frac{1/x}{-2/x^3} = \frac{1}{x} \cdot \frac{x^3}{-2} = -\frac{x^2}{2}$.
5. Evaluate the limit: as $x \to 0^+$, $-x^2/2 \to 0$.

Answer: 0

1. Identify the form: as $x \to \infty$, both terms go to infinity, so this is $\infty - \infty$.
2. Rationalize the expression by multiplying by the conjugate: $\frac{(\sqrt{x^2+x}-x)(\sqrt{x^2+x}+x)}{\sqrt{x^2+x}+x}$.
3. Simplify the numerator: $(\sqrt{x^2+x})^2 - x^2 = x^2 + x - x^2 = x$.
4. The limit is now $\lim_{x \to \infty} \frac{x}{\sqrt{x^2+x}+x}$, which is $\infty / \infty$.
5. Apply L'Hopital's rule: the derivative of the top is $1$. The derivative of the bottom is $\frac{2x+1}{2\sqrt{x^2+x}} + 1$.
6. As $x \to \infty$, the fraction $\frac{2x+1}{2\sqrt{x^2+x}}$ approaches $1$.
7. The limit becomes $1 / (1 + 1) = 1/2$.

Answer: 1/2

1. Identify the form: as $x \to 0^+$, the base $(1+x) \to 1$ and the exponent $1/x \to \infty$. This is a $1^\infty$ form.
2. Set $y = (1+x)^{1/x}$ and take the natural log: $\ln y = \ln((1+x)^{1/x}) = \frac{1}{x} \ln(1+x)$.
3. Rewrite as a quotient: $\lim_{x \to 0^+} \frac{\ln(1+x)}{x}$. This is a $0/0$ form.
4. Apply L'Hopital's rule: the derivative of $\ln(1+x)$ is $1/(1+x)$ and the derivative of $x$ is $1$.
5. Evaluate the limit: $\lim_{x \to 0^+} \frac{1/(1+x)}{1} = 1$.
6. Since $\ln y \to 1$, then $y \to e^1 = e$.

Answer: e

1. Calculate the average slope: $f(0) = 0$ and $f(2) = 2^3 - 2 = 6$. The average slope is $(6 - 0) / (2 - 0) = 3$.
2. Find the derivative: $f'(x) = 3x^2 - 1$.
3. Set the derivative equal to the average slope: $3c^2 - 1 = 3$.
4. Solve for $c$: $3c^2 = 4$, so $c^2 = 4/3$, which means $c = \pm 2/\sqrt{3}$.
5. Check which value is in the interval $[0, 2]$. Only $c = 2/\sqrt{3}$ is valid.

Answer: $c = \frac{2}{\sqrt{3}}$

1. Check continuity: $f(x)$ is continuous on $[-1, 1]$.
2. Check endpoints: $f(-1) = (-1)^{2/3} = 1$ and $f(1) = 1^{2/3} = 1$. Endpoints are equal.
3. Check differentiability: $f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$.
4. The derivative is undefined at $x = 0$, which is inside the open interval $(-1, 1)$.
5. Since the function is not differentiable on the open interval, Rolle's theorem does not apply.

Answer: No, because $f(x)$ is not differentiable at $x = 0$.

1. Calculate the average slope: $f(1) = \ln(1) = 0$ and $f(e) = \ln(e) = 1$. The average slope is $(1 - 0) / (e - 1) = 1 / (e - 1)$.
2. Find the derivative: $f'(x) = 1/x$.
3. Set the derivative equal to the average slope: $1/c = 1 / (e - 1)$.
4. Solve for $c$: $c = e - 1$.
5. Check if $c$ is in the interval $[1, e]$. Since $e \approx 2.718$, $c \approx 1.718$, which is between $1$ and $2.718$.

Answer: $c = e - 1$

1. Find the derivative: $f'(x) = 6x^2 - 6x - 12$.
2. Factor the derivative: $f'(x) = 6(x^2 - x - 2) = 6(x - 2)(x + 1)$.
3. Identify critical points where $f'(x) = 0$: $x = 2$ and $x = -1$.
4. Test intervals: For $x < -1$, $f'(-2) = 6(-4)(-1) = 24 > 0$ (Increasing). For $-1 < x < 2$, $f'(0) = -12 < 0$ (Decreasing). For $x > 2$, $f'(3) = 6(1)(4) = 24 > 0$ (Increasing).

Answer: Increasing on $(-\infty, -1) \cup (2, \infty)$, decreasing on $(-1, 2)$

1. Find the derivative: $f'(x) = e^x + 1$.
2. Analyze the range of the exponential function: $e^x$ is always greater than $0$ for all $x$.
3. Therefore, $f'(x) = e^x + 1$ must always be greater than $1$, which means $f'(x) > 0$ for all $x$.
4. By the Mean Value Theorem, since the derivative is always positive, the function is strictly increasing.

Answer: Since $f'(x) = e^x + 1 > 0$ for all $x$, the function is strictly increasing.

1. Set the derivative to be less than zero: $\frac{x-3}{x^2} < 0$.
2. Since $x^2$ is always positive for $x > 0$, the sign of the fraction depends only on the numerator.
3. Solve $x - 3 < 0$, which gives $x < 3$.
4. Combine this with the domain $x > 0$ to get the interval $(0, 3)$.

Answer: The function is decreasing on the interval $(0, 3)$

1. Find the derivative: $f'(x) = 3x^2 - 6x - 9$.
2. Set $f'(x) = 0$: $3(x^2 - 2x - 3) = 0$, which factors to $3(x-3)(x+1) = 0$. Critical points are $x = -1$ and $x = 3$.
3. Test intervals: For $x < -1$, $f'(-2) = 15 > 0$ (increasing). For $-1 < x < 3$, $f'(0) = -9 < 0$ (decreasing). For $x > 3$, $f'(4) = 15 > 0$ (increasing).
4. Classify: At $x = -1$, sign changes $(+) \to (-)$, so it is a local max. At $x = 3$, sign changes $(-) \to (+)$, so it is a local min.
5. Calculate values: $f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = 10$. $f(3) = (3)^3 - 3(3)^2 - 9(3) + 5 = -22$.

Answer: Local maximum of $10$ at $x = -1$; local minimum of $-22$ at $x = 3$.

1. Find the derivative: $f'(x) = 4x^3 - 8x$.
2. Set $f'(x) = 0$: $4x(x^2 - 2) = 0$. Critical points are $x = 0, x = \sqrt{2}, x = -\sqrt{2}$.
3. Test intervals: $x < -\sqrt{2}$ (e.g., $-2$): $f'(-2) = -32 + 16 = -16 < 0$ (decreasing).
4. Interval $-\sqrt{2} < x < 0$ (e.g., $-1$): $f'(-1) = -4 + 8 = 4 > 0$ (increasing).
5. Interval $0 < x < \sqrt{2}$ (e.g., $1$): $f'(1) = 4 - 8 = -4 < 0$ (decreasing).
6. Interval $x > \sqrt{2}$ (e.g., $2$): $f'(2) = 32 - 16 = 16 > 0$ (increasing).

Answer: Increasing on $(-\sqrt{2}, 0) \cup (\sqrt{2}, \infty)$; decreasing on $(-\infty, -\sqrt{2}) \cup (0, \sqrt{2})$.

1. Find the derivative: $f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$.
2. Identify critical points: $f'(x)$ is never zero, but it is undefined at $x = 0$. Since $x = 0$ is in the domain of $f$, it is a critical point.
3. Test intervals: For $x < 0$, $\sqrt[3]{x}$ is negative, so $f'(x) < 0$ (decreasing).
4. For $x > 0$, $\sqrt[3]{x}$ is positive, so $f'(x) > 0$ (increasing).
5. Classify: The sign changes from negative to positive at $x = 0$, so it is a local minimum.

Answer: Critical point at $x = 0$, which is a local minimum.