Chapter 5: Curve Sketching, L'Hôpital's Rule & Optimization
This chapter develops the analytical tools needed to understand the complete shape of a function's graph using calculus. While Chapter 4 introduced related rates, basic optimization, and the Mean Value Theorem, here we take a deeper and more systematic approach: we classify critical points, determine concavity, sketch curves from scratch, evaluate tricky limits with L'Hôpital's Rule, and solve sophisticated optimization problems. These topics form the backbone of the "Analytical Applications of Differentiation" unit on the AP exam and appear on virtually every administration of both the AB and BC tests.
Table of Contents
5.1Critical Points and Extreme Values
The starting point for understanding the shape of a graph is identifying where the function might change direction. These special locations are called critical points, and they are the only candidates for local (and absolute) extrema.
Definition: Critical Point
A number $c$ in the domain of $f$ is a critical point (or critical number) of $f$ if either:
A number $c$ in the domain of $f$ is a critical point (or critical number) of $f$ if either:
- $f'(c) = 0$, or
- $f'(c)$ does not exist.
The derivative equals zero at the top of a hill or the bottom of a valley, where the tangent line is horizontal. The derivative fails to exist at corners, cusps, or vertical tangent lines. Both situations require investigation.
Extreme Value Theorem
If $f$ is continuous on a closed interval $[a, b]$, then $f$ attains both an absolute maximum and an absolute minimum on $[a, b]$. These extreme values occur either at critical points in the interior $(a, b)$ or at the endpoints $a$ and $b$.
If $f$ is continuous on a closed interval $[a, b]$, then $f$ attains both an absolute maximum and an absolute minimum on $[a, b]$. These extreme values occur either at critical points in the interior $(a, b)$ or at the endpoints $a$ and $b$.
The Extreme Value Theorem gives us a clean algorithm for finding absolute extrema on a closed interval: find all critical points, evaluate $f$ at each critical point and at both endpoints, then compare.
Procedure: Finding Absolute Extrema on $[a, b]$
- Find $f'(x)$ and solve $f'(x) = 0$. Also identify where $f'(x)$ does not exist.
- List all critical points $c$ that lie in the open interval $(a, b)$.
- Evaluate $f$ at each critical point and at the endpoints $a$ and $b$.
- The largest value is the absolute maximum; the smallest is the absolute minimum.
Example 5.1.1: Find the absolute extrema of $f(x) = x^3 - 3x^2 + 1$ on the interval $[-1, 4]$.
Solution: Compute the derivative: $$f'(x) = 3x^2 - 6x = 3x(x - 2)$$ Setting $f'(x) = 0$: $3x(x - 2) = 0$ gives $x = 0$ and $x = 2$. Both lie in $(-1, 4)$, so both are critical points.
Evaluate $f$ at the critical points and endpoints:
Solution: Compute the derivative: $$f'(x) = 3x^2 - 6x = 3x(x - 2)$$ Setting $f'(x) = 0$: $3x(x - 2) = 0$ gives $x = 0$ and $x = 2$. Both lie in $(-1, 4)$, so both are critical points.
Evaluate $f$ at the critical points and endpoints:
- $f(-1) = (-1)^3 - 3(-1)^2 + 1 = -1 - 3 + 1 = -3$
- $f(0) = 0 - 0 + 1 = 1$
- $f(2) = 8 - 12 + 1 = -3$
- $f(4) = 64 - 48 + 1 = 17$
Example 5.1.2: Find the critical points of $f(x) = x^{2/3}(x - 5)$.
Solution: First expand: $f(x) = x^{5/3} - 5x^{2/3}$. Differentiate using the power rule: $$f'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5}{3}x^{-1/3}\!\left(x - 2\right) = \frac{5(x-2)}{3x^{1/3}}$$ Setting the numerator equal to zero: $5(x - 2) = 0$ gives $x = 2$. The denominator is zero when $x = 0$, so $f'(0)$ does not exist (but $f(0) = 0$ is defined).
The critical points are $x = 0$ (where $f'$ does not exist) and $x = 2$ (where $f' = 0$).
Solution: First expand: $f(x) = x^{5/3} - 5x^{2/3}$. Differentiate using the power rule: $$f'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5}{3}x^{-1/3}\!\left(x - 2\right) = \frac{5(x-2)}{3x^{1/3}}$$ Setting the numerator equal to zero: $5(x - 2) = 0$ gives $x = 2$. The denominator is zero when $x = 0$, so $f'(0)$ does not exist (but $f(0) = 0$ is defined).
The critical points are $x = 0$ (where $f'$ does not exist) and $x = 2$ (where $f' = 0$).
AP Tip: When finding critical points, do not forget to check where $f'(x)$ does not exist. A common mistake is to only solve $f'(x) = 0$ and miss cusps or corners. Also, a critical point must be in the domain of $f$ itself -- if $f$ is not defined at $c$, then $c$ is not a critical point.
5.2First Derivative Test
Once we have found the critical points, we need to classify each one. The first derivative test does this by examining the sign of $f'(x)$ on either side of the critical point. The logic is simple: if $f$ changes from increasing to decreasing at $c$, there is a local maximum at $c$; if $f$ changes from decreasing to increasing, there is a local minimum.
First Derivative Test
Let $c$ be a critical point of a continuous function $f$.
Let $c$ be a critical point of a continuous function $f$.
- If $f'(x)$ changes from positive to negative at $c$, then $f(c)$ is a local maximum.
- If $f'(x)$ changes from negative to positive at $c$, then $f(c)$ is a local minimum.
- If $f'(x)$ does not change sign at $c$ (positive on both sides or negative on both sides), then $f(c)$ is neither a local maximum nor a local minimum.
Sign Chart Method
To apply the first derivative test systematically:
To apply the first derivative test systematically:
- Find all critical points and mark them on a number line.
- Choose a test point in each interval between consecutive critical points.
- Evaluate $f'$ at each test point to determine the sign ($+$ or $-$).
- Record whether $f$ is increasing ($f' > 0$) or decreasing ($f' < 0$) on each interval.
- Classify each critical point based on the sign changes.
Example 5.2.1: Use the first derivative test to classify the critical points of $f(x) = 2x^3 - 9x^2 + 12x - 3$.
Solution: Compute the derivative: $$f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2)$$ Critical points: $x = 1$ and $x = 2$.
Build a sign chart. The factors are $(x - 1)$ and $(x - 2)$:
At $x = 2$: $f'$ changes from $-$ to $+$, so $f(2) = 16 - 36 + 24 - 3 = 1$ is a local minimum.
Solution: Compute the derivative: $$f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2)$$ Critical points: $x = 1$ and $x = 2$.
Build a sign chart. The factors are $(x - 1)$ and $(x - 2)$:
- For $x < 1$ (test $x = 0$): $f'(0) = 6(0-1)(0-2) = 6(-1)(-2) = 12 > 0$. $f$ is increasing.
- For $1 < x < 2$ (test $x = 1.5$): $f'(1.5) = 6(0.5)(-0.5) = -1.5 < 0$. $f$ is decreasing.
- For $x > 2$ (test $x = 3$): $f'(3) = 6(2)(1) = 12 > 0$. $f$ is increasing.
At $x = 2$: $f'$ changes from $-$ to $+$, so $f(2) = 16 - 36 + 24 - 3 = 1$ is a local minimum.
Example 5.2.2: Find the intervals on which $g(x) = x^4 - 4x^3$ is increasing and decreasing, and classify all critical points.
Solution: Compute the derivative: $$g'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$$ Setting $g'(x) = 0$: $4x^2(x - 3) = 0$ gives $x = 0$ and $x = 3$.
Sign chart:
At $x = 3$: $f'$ changes from $-$ to $+$, so $g(3) = 81 - 108 = -27$ is a local minimum.
$g$ is decreasing on $(-\infty, 3)$ and increasing on $(3, \infty)$.
Solution: Compute the derivative: $$g'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$$ Setting $g'(x) = 0$: $4x^2(x - 3) = 0$ gives $x = 0$ and $x = 3$.
Sign chart:
- For $x < 0$ (test $x = -1$): $g'(-1) = 4(1)(-4) = -16 < 0$. Decreasing.
- For $0 < x < 3$ (test $x = 1$): $g'(1) = 4(1)(-2) = -8 < 0$. Decreasing.
- For $x > 3$ (test $x = 4$): $g'(4) = 4(16)(1) = 64 > 0$. Increasing.
At $x = 3$: $f'$ changes from $-$ to $+$, so $g(3) = 81 - 108 = -27$ is a local minimum.
$g$ is decreasing on $(-\infty, 3)$ and increasing on $(3, \infty)$.
AP Tip: A common AP exam question provides a graph of $f'(x)$ (not $f(x)$) and asks you to determine where $f$ has local extrema. Look for sign changes: where $f'$ crosses the $x$-axis from positive to negative is a local max; from negative to positive is a local min. If $f'$ touches the axis but does not cross, there is no extremum.
5.3Second Derivative Test
The second derivative tells us about the concavity of a function, which describes whether the graph curves upward or downward. This information provides an alternative method for classifying critical points and is essential for a complete understanding of a function's shape.
Definition: Concavity
- A function $f$ is concave up on an interval if $f''(x) > 0$ on that interval. The graph lies above its tangent lines and curves upward (bowl shape).
- A function $f$ is concave down on an interval if $f''(x) < 0$ on that interval. The graph lies below its tangent lines and curves downward (dome shape).
Definition: Inflection Point
A point $(c, f(c))$ is an inflection point of $f$ if $f$ is continuous at $c$ and the concavity of $f$ changes at $c$ (from concave up to concave down, or vice versa). At an inflection point, $f''(c) = 0$ or $f''(c)$ does not exist. However, $f''(c) = 0$ alone is not sufficient -- the concavity must actually change.
A point $(c, f(c))$ is an inflection point of $f$ if $f$ is continuous at $c$ and the concavity of $f$ changes at $c$ (from concave up to concave down, or vice versa). At an inflection point, $f''(c) = 0$ or $f''(c)$ does not exist. However, $f''(c) = 0$ alone is not sufficient -- the concavity must actually change.
Second Derivative Test for Local Extrema
Suppose $f'(c) = 0$ and $f''$ is continuous near $c$.
Suppose $f'(c) = 0$ and $f''$ is continuous near $c$.
- If $f''(c) > 0$, then $f$ has a local minimum at $c$ (the graph is concave up, forming a valley).
- If $f''(c) < 0$, then $f$ has a local maximum at $c$ (the graph is concave down, forming a hill).
- If $f''(c) = 0$, the test is inconclusive. You must fall back to the first derivative test.
The second derivative test is often quicker than the first derivative test because it requires evaluating $f''$ at only one point rather than building a full sign chart. However, it fails when $f''(c) = 0$, which happens more often than students expect.
Example 5.3.1: Use the second derivative test to classify the critical points of $f(x) = x^3 - 6x^2 + 9x + 2$.
Solution: First derivative: $$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$$ Critical points: $x = 1$ and $x = 3$.
Second derivative: $$f''(x) = 6x - 12$$ Evaluate at each critical point:
Solution: First derivative: $$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$$ Critical points: $x = 1$ and $x = 3$.
Second derivative: $$f''(x) = 6x - 12$$ Evaluate at each critical point:
- $f''(1) = 6(1) - 12 = -6 < 0$. Concave down, so $f(1) = 1 - 6 + 9 + 2 = 6$ is a local maximum.
- $f''(3) = 6(3) - 12 = 6 > 0$. Concave up, so $f(3) = 27 - 54 + 27 + 2 = 2$ is a local minimum.
Example 5.3.2: Find the intervals of concavity and inflection points of $g(x) = x^4 - 6x^2 + 8x + 3$.
Solution: Compute derivatives: $$g'(x) = 4x^3 - 12x + 8$$ $$g''(x) = 12x^2 - 12 = 12(x^2 - 1) = 12(x - 1)(x + 1)$$ Set $g''(x) = 0$: $x = -1$ and $x = 1$.
Sign chart for $g''$:
$g(-1) = 1 - 6 - 8 + 3 = -10$. Inflection point at $(-1, -10)$.
$g(1) = 1 - 6 + 8 + 3 = 6$. Inflection point at $(1, 6)$.
$g$ is concave up on $(-\infty, -1) \cup (1, \infty)$ and concave down on $(-1, 1)$.
Solution: Compute derivatives: $$g'(x) = 4x^3 - 12x + 8$$ $$g''(x) = 12x^2 - 12 = 12(x^2 - 1) = 12(x - 1)(x + 1)$$ Set $g''(x) = 0$: $x = -1$ and $x = 1$.
Sign chart for $g''$:
- For $x < -1$ (test $x = -2$): $g''(-2) = 12(4-1) = 36 > 0$. Concave up.
- For $-1 < x < 1$ (test $x = 0$): $g''(0) = 12(0-1) = -12 < 0$. Concave down.
- For $x > 1$ (test $x = 2$): $g''(2) = 12(4-1) = 36 > 0$. Concave up.
$g(-1) = 1 - 6 - 8 + 3 = -10$. Inflection point at $(-1, -10)$.
$g(1) = 1 - 6 + 8 + 3 = 6$. Inflection point at $(1, 6)$.
$g$ is concave up on $(-\infty, -1) \cup (1, \infty)$ and concave down on $(-1, 1)$.
AP Tip: The second derivative test is inconclusive when $f''(c) = 0$, but this does not mean $c$ is an inflection point. For example, $f(x) = x^4$ has $f'(0) = 0$ and $f''(0) = 0$, but $x = 0$ is a local minimum (not an inflection point). Always verify that concavity actually changes before declaring an inflection point.
5.4Complete Curve Sketching
With the first and second derivative tests in hand, we can now develop a systematic procedure for sketching the graph of a function without a graphing calculator. This is one of the most important skills tested on the AP exam, as it demonstrates a deep understanding of how calculus reveals the shape of a curve.
Systematic Curve Sketching Procedure
Given a function $f(x)$, analyze the following in order:
Given a function $f(x)$, analyze the following in order:
- Domain: Determine where $f$ is defined.
- Symmetry: Check if $f$ is even ($f(-x) = f(x)$), odd ($f(-x) = -f(x)$), or neither.
- Intercepts: Find $y$-intercept ($x = 0$) and $x$-intercepts ($f(x) = 0$).
- Asymptotes: Vertical (denominator = 0), horizontal ($\lim_{x \to \pm\infty}$), oblique (degree of numerator = degree of denominator + 1).
- First derivative analysis: Find $f'(x)$, locate critical points, determine increasing/decreasing intervals and local extrema.
- Second derivative analysis: Find $f''(x)$, determine concavity intervals and inflection points.
- Key points: Plot all critical points, inflection points, intercepts, and a few additional points if needed.
- Sketch: Connect the points, respecting all the information gathered above.
Use the interactive graph below to explore how the first and second derivatives reveal the shape of a polynomial. Adjust the sliders $a$, $b$, $c$, and $d$ to change the cubic $f(x) = ax^3 + bx^2 + cx + d$, and observe how $f'(x)$ and $f''(x)$ respond.
Explore how the first and second derivatives reveal the shape of a curve. Blue solid: $f(x)$, red dashed: $f'(x)$, green dotted: $f''(x)$.
Example 5.4.1 (Rational Function): Sketch the graph of $f(x) = \dfrac{2x^2}{x^2 - 1}$.
Solution:
1. Domain: $x^2 - 1 = 0$ when $x = \pm 1$. Domain: $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$.
2. Symmetry: $f(-x) = \frac{2(-x)^2}{(-x)^2 - 1} = \frac{2x^2}{x^2 - 1} = f(x)$. The function is even.
3. Intercepts: $y$-intercept: $f(0) = 0$. $x$-intercept: $2x^2 = 0 \Rightarrow x = 0$. The only intercept is the origin $(0, 0)$.
4. Asymptotes:
Vertical: $x = -1$ and $x = 1$.
Horizontal: $\lim_{x \to \pm\infty} \frac{2x^2}{x^2 - 1} = \lim_{x \to \pm\infty} \frac{2}{1 - 1/x^2} = 2$. Horizontal asymptote: $y = 2$.
5. First derivative: $$f'(x) = \frac{4x(x^2-1) - 2x^2 \cdot 2x}{(x^2-1)^2} = \frac{4x^3 - 4x - 4x^3}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2}$$ $f'(x) = 0$ when $x = 0$. Sign analysis: $f'(x) > 0$ for $x < 0$ (in the domain) and $f'(x) < 0$ for $x > 0$ (in the domain). So $f$ is increasing on $(-\infty, -1)$ and $(-1, 0)$, decreasing on $(0, 1)$ and $(1, \infty)$. Local maximum at $(0, 0)$.
6. Second derivative: By the quotient rule (computation omitted for brevity): $$f''(x) = \frac{4(3x^2 + 1)}{(x^2 - 1)^3}$$ The numerator $4(3x^2 + 1) > 0$ always. The sign depends on $(x^2 - 1)^3$: positive when $|x| > 1$, negative when $|x| < 1$. So $f$ is concave up on $(-\infty, -1) \cup (1, \infty)$ and concave down on $(-1, 1)$. No inflection points (concavity changes at the asymptotes, not at points on the graph).
7. Sketch: The graph passes through the origin with a local maximum, approaches $y = 2$ from below as $x \to \pm\infty$, and has vertical asymptotes at $x = \pm 1$.
Solution:
1. Domain: $x^2 - 1 = 0$ when $x = \pm 1$. Domain: $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$.
2. Symmetry: $f(-x) = \frac{2(-x)^2}{(-x)^2 - 1} = \frac{2x^2}{x^2 - 1} = f(x)$. The function is even.
3. Intercepts: $y$-intercept: $f(0) = 0$. $x$-intercept: $2x^2 = 0 \Rightarrow x = 0$. The only intercept is the origin $(0, 0)$.
4. Asymptotes:
Vertical: $x = -1$ and $x = 1$.
Horizontal: $\lim_{x \to \pm\infty} \frac{2x^2}{x^2 - 1} = \lim_{x \to \pm\infty} \frac{2}{1 - 1/x^2} = 2$. Horizontal asymptote: $y = 2$.
5. First derivative: $$f'(x) = \frac{4x(x^2-1) - 2x^2 \cdot 2x}{(x^2-1)^2} = \frac{4x^3 - 4x - 4x^3}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2}$$ $f'(x) = 0$ when $x = 0$. Sign analysis: $f'(x) > 0$ for $x < 0$ (in the domain) and $f'(x) < 0$ for $x > 0$ (in the domain). So $f$ is increasing on $(-\infty, -1)$ and $(-1, 0)$, decreasing on $(0, 1)$ and $(1, \infty)$. Local maximum at $(0, 0)$.
6. Second derivative: By the quotient rule (computation omitted for brevity): $$f''(x) = \frac{4(3x^2 + 1)}{(x^2 - 1)^3}$$ The numerator $4(3x^2 + 1) > 0$ always. The sign depends on $(x^2 - 1)^3$: positive when $|x| > 1$, negative when $|x| < 1$. So $f$ is concave up on $(-\infty, -1) \cup (1, \infty)$ and concave down on $(-1, 1)$. No inflection points (concavity changes at the asymptotes, not at points on the graph).
7. Sketch: The graph passes through the origin with a local maximum, approaches $y = 2$ from below as $x \to \pm\infty$, and has vertical asymptotes at $x = \pm 1$.
Example 5.4.2 (Polynomial): Sketch the graph of $f(x) = x^4 - 4x^3 + 10$.
Solution:
1. Domain: All real numbers.
2. Symmetry: Neither even nor odd.
3. Intercepts: $y$-intercept: $f(0) = 10$. $x$-intercepts are difficult to find algebraically, so we skip them for now.
4. Asymptotes: None (polynomial).
5. First derivative: $$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$$ Critical points: $x = 0$ and $x = 3$.
Sign chart: $f'(x) < 0$ for $x < 0$, $f'(x) < 0$ for $0 < x < 3$, $f'(x) > 0$ for $x > 3$.
At $x = 0$: no sign change, so neither max nor min.
At $x = 3$: sign changes from $-$ to $+$, so local minimum at $f(3) = 81 - 108 + 10 = -17$.
6. Second derivative: $$f''(x) = 12x^2 - 24x = 12x(x - 2)$$ $f''(x) = 0$ at $x = 0$ and $x = 2$.
Sign chart: $f'' > 0$ for $x < 0$, $f'' < 0$ for $0 < x < 2$, $f'' > 0$ for $x > 2$.
Inflection points at $(0, 10)$ and $(2, f(2)) = (2, 16 - 32 + 10) = (2, -6)$.
7. Sketch: The graph starts high on the left (concave up), passes through the inflection point $(0, 10)$, becomes concave down, passes through another inflection point at $(2, -6)$, then reaches a local minimum at $(3, -17)$ before increasing to the right.
Solution:
1. Domain: All real numbers.
2. Symmetry: Neither even nor odd.
3. Intercepts: $y$-intercept: $f(0) = 10$. $x$-intercepts are difficult to find algebraically, so we skip them for now.
4. Asymptotes: None (polynomial).
5. First derivative: $$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$$ Critical points: $x = 0$ and $x = 3$.
Sign chart: $f'(x) < 0$ for $x < 0$, $f'(x) < 0$ for $0 < x < 3$, $f'(x) > 0$ for $x > 3$.
At $x = 0$: no sign change, so neither max nor min.
At $x = 3$: sign changes from $-$ to $+$, so local minimum at $f(3) = 81 - 108 + 10 = -17$.
6. Second derivative: $$f''(x) = 12x^2 - 24x = 12x(x - 2)$$ $f''(x) = 0$ at $x = 0$ and $x = 2$.
Sign chart: $f'' > 0$ for $x < 0$, $f'' < 0$ for $0 < x < 2$, $f'' > 0$ for $x > 2$.
Inflection points at $(0, 10)$ and $(2, f(2)) = (2, 16 - 32 + 10) = (2, -6)$.
7. Sketch: The graph starts high on the left (concave up), passes through the inflection point $(0, 10)$, becomes concave down, passes through another inflection point at $(2, -6)$, then reaches a local minimum at $(3, -17)$ before increasing to the right.
Example 5.4.3 (Exponential): Sketch the graph of $f(x) = xe^{-x}$.
Solution:
1. Domain: All real numbers.
2. Symmetry: Neither even nor odd.
3. Intercepts: $f(0) = 0$. The only $x$-intercept is $x = 0$.
4. End behavior: $\lim_{x \to \infty} xe^{-x} = 0$ (exponential decay dominates). $\lim_{x \to -\infty} xe^{-x} = -\infty$. Horizontal asymptote: $y = 0$ as $x \to \infty$.
5. First derivative: Using the product rule: $$f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$$ Since $e^{-x} > 0$ always, $f'(x) = 0$ when $1 - x = 0$, i.e., $x = 1$.
$f'(x) > 0$ for $x < 1$ (increasing) and $f'(x) < 0$ for $x > 1$ (decreasing).
Local maximum at $x = 1$: $f(1) = 1 \cdot e^{-1} = \frac{1}{e} \approx 0.368$.
6. Second derivative: $$f''(x) = -e^{-x}(1 - x) + e^{-x}(-1) = e^{-x}(x - 2)$$ $f''(x) = 0$ when $x = 2$. Since $e^{-x} > 0$, $f''$ changes sign at $x = 2$: concave down for $x < 2$ and concave up for $x > 2$.
Inflection point at $(2, 2e^{-2}) \approx (2, 0.271)$.
7. Sketch: The graph rises from $-\infty$ through the origin, reaches a local max at $(1, 1/e)$, then decreases toward the horizontal asymptote $y = 0$, with an inflection point at $x = 2$.
Solution:
1. Domain: All real numbers.
2. Symmetry: Neither even nor odd.
3. Intercepts: $f(0) = 0$. The only $x$-intercept is $x = 0$.
4. End behavior: $\lim_{x \to \infty} xe^{-x} = 0$ (exponential decay dominates). $\lim_{x \to -\infty} xe^{-x} = -\infty$. Horizontal asymptote: $y = 0$ as $x \to \infty$.
5. First derivative: Using the product rule: $$f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$$ Since $e^{-x} > 0$ always, $f'(x) = 0$ when $1 - x = 0$, i.e., $x = 1$.
$f'(x) > 0$ for $x < 1$ (increasing) and $f'(x) < 0$ for $x > 1$ (decreasing).
Local maximum at $x = 1$: $f(1) = 1 \cdot e^{-1} = \frac{1}{e} \approx 0.368$.
6. Second derivative: $$f''(x) = -e^{-x}(1 - x) + e^{-x}(-1) = e^{-x}(x - 2)$$ $f''(x) = 0$ when $x = 2$. Since $e^{-x} > 0$, $f''$ changes sign at $x = 2$: concave down for $x < 2$ and concave up for $x > 2$.
Inflection point at $(2, 2e^{-2}) \approx (2, 0.271)$.
7. Sketch: The graph rises from $-\infty$ through the origin, reaches a local max at $(1, 1/e)$, then decreases toward the horizontal asymptote $y = 0$, with an inflection point at $x = 2$.
5.5L'Hôpital's Rule
Many limits in calculus produce indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ when you try direct substitution. L'Hôpital's Rule provides an elegant technique for evaluating such limits by replacing the original limit with a limit of the ratio of derivatives.
L'Hôpital's Rule
Suppose $f$ and $g$ are differentiable on an open interval containing $a$ (except possibly at $a$ itself), and $g'(x) \neq 0$ near $a$. If $$\lim_{x \to a} f(x) = 0 \text{ and } \lim_{x \to a} g(x) = 0 \quad \left(\frac{0}{0} \text{ form}\right)$$ or $$\lim_{x \to a} |f(x)| = \infty \text{ and } \lim_{x \to a} |g(x)| = \infty \quad \left(\frac{\infty}{\infty} \text{ form}\right)$$ then $$\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}$$ provided the limit on the right side exists (or is $\pm\infty$). The same rule applies for $x \to a^+$, $x \to a^-$, $x \to \infty$, and $x \to -\infty$.
Suppose $f$ and $g$ are differentiable on an open interval containing $a$ (except possibly at $a$ itself), and $g'(x) \neq 0$ near $a$. If $$\lim_{x \to a} f(x) = 0 \text{ and } \lim_{x \to a} g(x) = 0 \quad \left(\frac{0}{0} \text{ form}\right)$$ or $$\lim_{x \to a} |f(x)| = \infty \text{ and } \lim_{x \to a} |g(x)| = \infty \quad \left(\frac{\infty}{\infty} \text{ form}\right)$$ then $$\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}$$ provided the limit on the right side exists (or is $\pm\infty$). The same rule applies for $x \to a^+$, $x \to a^-$, $x \to \infty$, and $x \to -\infty$.
When NOT to Use L'Hôpital's Rule:
- Not an indeterminate form: If direct substitution gives a definite value like $\frac{5}{0}$, $\frac{3}{7}$, or $\frac{0}{5}$, do NOT apply L'Hôpital's Rule. These are not indeterminate.
- Circular reasoning: The limit $\lim_{x \to 0}\frac{\sin x}{x}$ can be evaluated by L'Hôpital's Rule, but the derivative of $\sin x$ is typically derived using this very limit, making the argument circular.
- Algebraic simplification is easier: Before reaching for L'Hôpital's Rule, check if factoring, rationalizing, or simplifying resolves the indeterminate form more efficiently.
Other Indeterminate Forms
L'Hôpital's Rule directly handles only $\frac{0}{0}$ and $\frac{\infty}{\infty}$. Other indeterminate forms must be converted first:
- $0 \cdot \infty$: Rewrite as $\frac{0}{1/\infty} = \frac{0}{0}$ or $\frac{\infty}{1/0} = \frac{\infty}{\infty}$.
- $\infty - \infty$: Combine fractions or factor to create $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
- $0^0$, $1^\infty$, $\infty^0$: Take the natural logarithm. If $L = \lim f(x)^{g(x)}$, compute $\ln L = \lim g(x) \ln f(x)$, which is typically a $0 \cdot \infty$ form, then exponentiate: $L = e^{\ln L}$.
Example 5.5.1 (Basic Application): Evaluate $\displaystyle\lim_{x \to 0}\frac{e^x - 1}{x}$.
Solution: Direct substitution gives $\frac{e^0 - 1}{0} = \frac{0}{0}$, an indeterminate form. Apply L'Hôpital's Rule: $$\lim_{x \to 0}\frac{e^x - 1}{x} = \lim_{x \to 0}\frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(x)} = \lim_{x \to 0}\frac{e^x}{1} = e^0 = 1$$
Solution: Direct substitution gives $\frac{e^0 - 1}{0} = \frac{0}{0}$, an indeterminate form. Apply L'Hôpital's Rule: $$\lim_{x \to 0}\frac{e^x - 1}{x} = \lim_{x \to 0}\frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(x)} = \lim_{x \to 0}\frac{e^x}{1} = e^0 = 1$$
Example 5.5.2 (Repeated Application): Evaluate $\displaystyle\lim_{x \to 0}\frac{x - \sin x}{x^3}$.
Solution: Direct substitution gives $\frac{0 - 0}{0} = \frac{0}{0}$. Apply L'Hôpital's Rule: $$\lim_{x \to 0}\frac{1 - \cos x}{3x^2}$$ This is still $\frac{0}{0}$. Apply L'Hôpital's Rule again: $$\lim_{x \to 0}\frac{\sin x}{6x}$$ Still $\frac{0}{0}$. Apply one more time: $$\lim_{x \to 0}\frac{\cos x}{6} = \frac{1}{6}$$
Solution: Direct substitution gives $\frac{0 - 0}{0} = \frac{0}{0}$. Apply L'Hôpital's Rule: $$\lim_{x \to 0}\frac{1 - \cos x}{3x^2}$$ This is still $\frac{0}{0}$. Apply L'Hôpital's Rule again: $$\lim_{x \to 0}\frac{\sin x}{6x}$$ Still $\frac{0}{0}$. Apply one more time: $$\lim_{x \to 0}\frac{\cos x}{6} = \frac{1}{6}$$
Example 5.5.3 (Exponential Indeterminate Form): Evaluate $\displaystyle\lim_{x \to 0^+} x^x$.
Solution: This is the indeterminate form $0^0$. Let $L = \lim_{x \to 0^+} x^x$. Take the natural log: $$\ln L = \lim_{x \to 0^+} x \ln x$$ This is a $0 \cdot (-\infty)$ form. Rewrite as: $$\ln L = \lim_{x \to 0^+} \frac{\ln x}{1/x}$$ This is $\frac{-\infty}{\infty}$. Apply L'Hôpital's Rule: $$\ln L = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \frac{x^2}{-x} \cdot \frac{1}{1} = \lim_{x \to 0^+}(-x) = 0$$ Therefore $L = e^0 = 1$.
Solution: This is the indeterminate form $0^0$. Let $L = \lim_{x \to 0^+} x^x$. Take the natural log: $$\ln L = \lim_{x \to 0^+} x \ln x$$ This is a $0 \cdot (-\infty)$ form. Rewrite as: $$\ln L = \lim_{x \to 0^+} \frac{\ln x}{1/x}$$ This is $\frac{-\infty}{\infty}$. Apply L'Hôpital's Rule: $$\ln L = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \frac{x^2}{-x} \cdot \frac{1}{1} = \lim_{x \to 0^+}(-x) = 0$$ Therefore $L = e^0 = 1$.
AP Tip: On the AP exam, L'Hôpital's Rule appears primarily in the multiple-choice section. The most common mistake is applying it when the form is not actually indeterminate. Always verify that you have $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before differentiating. Also, differentiate the numerator and denominator separately -- do not use the quotient rule on the fraction.
5.6Optimization Revisited
In Chapter 4, we introduced optimization problems involving finding maximum or minimum values. Here we revisit the topic with a more systematic strategy and more challenging problems. Optimization is one of the most heavily tested application topics on the AP exam, appearing almost every year on the free-response section.
Systematic Optimization Strategy
- Draw a diagram and label all quantities with variables.
- Identify the objective function: Write the quantity to be maximized or minimized as a function of one or more variables.
- Establish the constraint: Use given relationships to express the objective function in terms of a single variable.
- Determine the domain: What are the physically meaningful values of the variable?
- Find critical points: Differentiate, set equal to zero, and solve.
- Classify: Use the first or second derivative test to confirm whether the critical point is a max or min.
- Answer the question: Return to the original context and state the answer with units.
The interactive graph below illustrates the classic open-box optimization problem. A rectangular sheet of dimensions $L \times W$ has squares of side $x$ cut from each corner, and the sides are folded up to form an open box. Adjust $L$ and $W$ to see how the volume function $V(x) = x(L - 2x)(W - 2x)$ changes and where the maximum occurs.
Adjust the sheet dimensions $L$ and $W$ and see how the optimal cut size changes. The red dot marks the maximum volume.
Example 5.6.1 (Maximize Volume): An open-top box is made from a $24 \times 24$ cm sheet of cardboard by cutting squares of side length $x$ from each corner and folding up the sides. Find the value of $x$ that maximizes the volume.
Solution:
Objective function: After cutting, the base is $(24 - 2x) \times (24 - 2x)$ and the height is $x$: $$V(x) = x(24 - 2x)^2$$ Domain: $0 < x < 12$ (each side must remain positive).
Expand and differentiate: $$V(x) = x(576 - 96x + 4x^2) = 4x^3 - 96x^2 + 576x$$ $$V'(x) = 12x^2 - 192x + 576 = 12(x^2 - 16x + 48) = 12(x - 4)(x - 12)$$ Setting $V'(x) = 0$: $x = 4$ or $x = 12$. Since $x = 12$ is an endpoint where $V = 0$, the only interior critical point is $x = 4$.
Classify: $V''(x) = 24x - 192$. At $x = 4$: $V''(4) = 96 - 192 = -96 < 0$. Concave down, confirming a local maximum.
Answer: The maximum volume is $V(4) = 4(24 - 8)^2 = 4(16)^2 = 4 \cdot 256 = 1024$ cm$^3$, achieved with a cut size of $x = 4$ cm.
Solution:
Objective function: After cutting, the base is $(24 - 2x) \times (24 - 2x)$ and the height is $x$: $$V(x) = x(24 - 2x)^2$$ Domain: $0 < x < 12$ (each side must remain positive).
Expand and differentiate: $$V(x) = x(576 - 96x + 4x^2) = 4x^3 - 96x^2 + 576x$$ $$V'(x) = 12x^2 - 192x + 576 = 12(x^2 - 16x + 48) = 12(x - 4)(x - 12)$$ Setting $V'(x) = 0$: $x = 4$ or $x = 12$. Since $x = 12$ is an endpoint where $V = 0$, the only interior critical point is $x = 4$.
Classify: $V''(x) = 24x - 192$. At $x = 4$: $V''(4) = 96 - 192 = -96 < 0$. Concave down, confirming a local maximum.
Answer: The maximum volume is $V(4) = 4(24 - 8)^2 = 4(16)^2 = 4 \cdot 256 = 1024$ cm$^3$, achieved with a cut size of $x = 4$ cm.
Example 5.6.2 (Minimize Cost): A manufacturer needs to produce cylindrical cans with a volume of $1000$ cm$^3$. The material for the top and bottom costs \$0.06 per cm$^2$, and the material for the side costs \$0.03 per cm$^2$. Find the dimensions that minimize the total cost.
Solution:
Variables: Let $r$ = radius and $h$ = height of the cylinder.
Constraint: $\pi r^2 h = 1000$, so $h = \frac{1000}{\pi r^2}$.
Objective function (cost): The top and bottom have combined area $2\pi r^2$. The lateral surface has area $2\pi r h$. $$C(r) = 0.06 \cdot 2\pi r^2 + 0.03 \cdot 2\pi r h = 0.12\pi r^2 + 0.06\pi r \cdot \frac{1000}{\pi r^2}$$ $$C(r) = 0.12\pi r^2 + \frac{60}{r}$$ Domain: $r > 0$.
Differentiate: $$C'(r) = 0.24\pi r - \frac{60}{r^2}$$ Set $C'(r) = 0$: $$0.24\pi r = \frac{60}{r^2} \implies r^3 = \frac{60}{0.24\pi} = \frac{250}{\pi}$$ $$r = \left(\frac{250}{\pi}\right)^{1/3} \approx 4.30 \text{ cm}$$ Classify: $C''(r) = 0.24\pi + \frac{120}{r^3}$. Since both terms are positive for $r > 0$, we have $C''(r) > 0$, confirming a minimum.
Find height: $$h = \frac{1000}{\pi r^2} = \frac{1000}{\pi\left(\frac{250}{\pi}\right)^{2/3}}$$ Numerically, $h \approx 17.21$ cm.
Answer: The cost is minimized when $r \approx 4.30$ cm and $h \approx 17.21$ cm. Note that $h \approx 4r$, meaning the height is about four times the radius -- the can is tall and narrow because the side material is cheaper.
Solution:
Variables: Let $r$ = radius and $h$ = height of the cylinder.
Constraint: $\pi r^2 h = 1000$, so $h = \frac{1000}{\pi r^2}$.
Objective function (cost): The top and bottom have combined area $2\pi r^2$. The lateral surface has area $2\pi r h$. $$C(r) = 0.06 \cdot 2\pi r^2 + 0.03 \cdot 2\pi r h = 0.12\pi r^2 + 0.06\pi r \cdot \frac{1000}{\pi r^2}$$ $$C(r) = 0.12\pi r^2 + \frac{60}{r}$$ Domain: $r > 0$.
Differentiate: $$C'(r) = 0.24\pi r - \frac{60}{r^2}$$ Set $C'(r) = 0$: $$0.24\pi r = \frac{60}{r^2} \implies r^3 = \frac{60}{0.24\pi} = \frac{250}{\pi}$$ $$r = \left(\frac{250}{\pi}\right)^{1/3} \approx 4.30 \text{ cm}$$ Classify: $C''(r) = 0.24\pi + \frac{120}{r^3}$. Since both terms are positive for $r > 0$, we have $C''(r) > 0$, confirming a minimum.
Find height: $$h = \frac{1000}{\pi r^2} = \frac{1000}{\pi\left(\frac{250}{\pi}\right)^{2/3}}$$ Numerically, $h \approx 17.21$ cm.
Answer: The cost is minimized when $r \approx 4.30$ cm and $h \approx 17.21$ cm. Note that $h \approx 4r$, meaning the height is about four times the radius -- the can is tall and narrow because the side material is cheaper.
AP Tip: In optimization problems, the AP exam often provides the constraint equation and asks you to justify that your critical point is indeed a maximum or minimum. Stating "the second derivative is negative" or "the first derivative changes from positive to negative" is sufficient. Simply finding the critical point without justification will not earn full credit.
5.7Practice Problems
Test your mastery of this chapter with the following twelve problems. Each problem targets a specific skill. Try to solve each one completely before expanding the solution.
Problem 1. Find all critical points of $f(x) = 3x^4 - 4x^3 - 12x^2 + 5$.
Show Solution
Differentiate:
$$f'(x) = 12x^3 - 12x^2 - 24x = 12x(x^2 - x - 2) = 12x(x - 2)(x + 1)$$
Setting $f'(x) = 0$: $x = 0$, $x = 2$, $x = -1$.
All three are in the domain, so the critical points are $x = -1$, $x = 0$, and $x = 2$.
All three are in the domain, so the critical points are $x = -1$, $x = 0$, and $x = 2$.
Problem 2. Use the first derivative test to classify the critical points of $f(x) = x^3 - \frac{3}{2}x^2 - 18x + 4$.
Show Solution
$f'(x) = 3x^2 - 3x - 18 = 3(x^2 - x - 6) = 3(x - 3)(x + 2)$
Critical points: $x = -2$ and $x = 3$.
Sign chart:
At $x = 3$: $f'$ changes from $-$ to $+$. Local minimum at $f(3) = 27 - \frac{27}{2} - 54 + 4 = -\frac{59}{2}$.
Critical points: $x = -2$ and $x = 3$.
Sign chart:
- $x < -2$: $f'(-3) = 3(-6)(1) > 0$. Increasing.
- $-2 < x < 3$: $f'(0) = 3(-3)(2) < 0$. Decreasing.
- $x > 3$: $f'(4) = 3(1)(6) > 0$. Increasing.
At $x = 3$: $f'$ changes from $-$ to $+$. Local minimum at $f(3) = 27 - \frac{27}{2} - 54 + 4 = -\frac{59}{2}$.
Problem 3. Find the intervals of concavity and all inflection points of $f(x) = x^4 - 4x^3 + 6x^2 - 4x$.
Show Solution
$f'(x) = 4x^3 - 12x^2 + 12x - 4 = 4(x^3 - 3x^2 + 3x - 1) = 4(x - 1)^3$
$f''(x) = 12(x - 1)^2$
$f''(x) = 0$ when $x = 1$. However, $f''(x) = 12(x-1)^2 \geq 0$ for all $x$, and $f''(x) > 0$ for all $x \neq 1$.
Since $f''$ does not change sign at $x = 1$ (it is non-negative on both sides), there is no inflection point. The function is concave up on $(-\infty, 1) \cup (1, \infty)$ and satisfies $f''(1) = 0$ without a concavity change.
$f''(x) = 12(x - 1)^2$
$f''(x) = 0$ when $x = 1$. However, $f''(x) = 12(x-1)^2 \geq 0$ for all $x$, and $f''(x) > 0$ for all $x \neq 1$.
Since $f''$ does not change sign at $x = 1$ (it is non-negative on both sides), there is no inflection point. The function is concave up on $(-\infty, 1) \cup (1, \infty)$ and satisfies $f''(1) = 0$ without a concavity change.
Problem 4. Evaluate $\displaystyle\lim_{x \to 1}\frac{x^3 - 1}{x^2 - 1}$ using L'Hôpital's Rule.
Show Solution
Direct substitution: $\frac{1 - 1}{1 - 1} = \frac{0}{0}$. This is an indeterminate form.
Apply L'Hôpital's Rule: $$\lim_{x \to 1}\frac{3x^2}{2x} = \frac{3(1)}{2(1)} = \frac{3}{2}$$ (Note: This could also be solved by factoring: $\frac{(x-1)(x^2+x+1)}{(x-1)(x+1)} = \frac{x^2+x+1}{x+1} \to \frac{3}{2}$.)
Apply L'Hôpital's Rule: $$\lim_{x \to 1}\frac{3x^2}{2x} = \frac{3(1)}{2(1)} = \frac{3}{2}$$ (Note: This could also be solved by factoring: $\frac{(x-1)(x^2+x+1)}{(x-1)(x+1)} = \frac{x^2+x+1}{x+1} \to \frac{3}{2}$.)
Problem 5. Evaluate $\displaystyle\lim_{x \to 0}\frac{e^x - 1 - x}{x^2}$.
Show Solution
Direct substitution: $\frac{1 - 1 - 0}{0} = \frac{0}{0}$.
First application of L'Hôpital's Rule: $$\lim_{x \to 0}\frac{e^x - 1}{2x} = \frac{0}{0}$$ Second application: $$\lim_{x \to 0}\frac{e^x}{2} = \frac{1}{2}$$
First application of L'Hôpital's Rule: $$\lim_{x \to 0}\frac{e^x - 1}{2x} = \frac{0}{0}$$ Second application: $$\lim_{x \to 0}\frac{e^x}{2} = \frac{1}{2}$$
Problem 6. Evaluate $\displaystyle\lim_{x \to \infty}\left(1 + \frac{3}{x}\right)^x$.
Show Solution
This is the indeterminate form $1^\infty$. Let $L = \lim_{x \to \infty}\left(1 + \frac{3}{x}\right)^x$.
Take the natural log: $$\ln L = \lim_{x \to \infty} x \ln\!\left(1 + \frac{3}{x}\right)$$ This is $\infty \cdot 0$. Rewrite: $$\ln L = \lim_{x \to \infty}\frac{\ln\!\left(1 + \frac{3}{x}\right)}{1/x}$$ This is $\frac{0}{0}$. Let $t = 1/x$, so as $x \to \infty$, $t \to 0^+$: $$\ln L = \lim_{t \to 0^+}\frac{\ln(1 + 3t)}{t}$$ Apply L'Hôpital's Rule: $$\ln L = \lim_{t \to 0^+}\frac{\frac{3}{1+3t}}{1} = \frac{3}{1} = 3$$ Therefore $L = e^3$.
Take the natural log: $$\ln L = \lim_{x \to \infty} x \ln\!\left(1 + \frac{3}{x}\right)$$ This is $\infty \cdot 0$. Rewrite: $$\ln L = \lim_{x \to \infty}\frac{\ln\!\left(1 + \frac{3}{x}\right)}{1/x}$$ This is $\frac{0}{0}$. Let $t = 1/x$, so as $x \to \infty$, $t \to 0^+$: $$\ln L = \lim_{t \to 0^+}\frac{\ln(1 + 3t)}{t}$$ Apply L'Hôpital's Rule: $$\ln L = \lim_{t \to 0^+}\frac{\frac{3}{1+3t}}{1} = \frac{3}{1} = 3$$ Therefore $L = e^3$.
Problem 7. Sketch the graph of $f(x) = \dfrac{x}{x^2 + 1}$ by finding all relevant features (intercepts, symmetry, asymptotes, extrema, concavity, inflection points).
Show Solution
Domain: All real numbers ($x^2 + 1 > 0$ always).
Symmetry: $f(-x) = \frac{-x}{x^2+1} = -f(x)$, so $f$ is odd.
Intercepts: $f(0) = 0$. Only intercept is the origin.
Asymptotes: No vertical asymptotes. $\lim_{x \to \pm\infty}\frac{x}{x^2+1} = 0$. Horizontal asymptote: $y = 0$.
First derivative: $$f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$$ $f'(x) = 0$ when $1 - x^2 = 0$, so $x = \pm 1$.
$f'(x) > 0$ on $(-1, 1)$ and $f'(x) < 0$ on $(-\infty, -1) \cup (1, \infty)$.
Local min at $(-1, -1/2)$ and local max at $(1, 1/2)$.
Second derivative: (by quotient rule) $$f''(x) = \frac{2x(x^2 - 3)}{(x^2+1)^3}$$ $f''(x) = 0$ when $x = 0$, $x = \sqrt{3}$, or $x = -\sqrt{3}$.
Three inflection points: $(0, 0)$, $(\sqrt{3}, \frac{\sqrt{3}}{4})$, $(-\sqrt{3}, -\frac{\sqrt{3}}{4})$.
Sketch: An odd function passing through the origin, rising to a local max of $1/2$ at $x = 1$, then decaying to $0$; symmetric behavior on the left with a local min of $-1/2$ at $x = -1$.
Symmetry: $f(-x) = \frac{-x}{x^2+1} = -f(x)$, so $f$ is odd.
Intercepts: $f(0) = 0$. Only intercept is the origin.
Asymptotes: No vertical asymptotes. $\lim_{x \to \pm\infty}\frac{x}{x^2+1} = 0$. Horizontal asymptote: $y = 0$.
First derivative: $$f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$$ $f'(x) = 0$ when $1 - x^2 = 0$, so $x = \pm 1$.
$f'(x) > 0$ on $(-1, 1)$ and $f'(x) < 0$ on $(-\infty, -1) \cup (1, \infty)$.
Local min at $(-1, -1/2)$ and local max at $(1, 1/2)$.
Second derivative: (by quotient rule) $$f''(x) = \frac{2x(x^2 - 3)}{(x^2+1)^3}$$ $f''(x) = 0$ when $x = 0$, $x = \sqrt{3}$, or $x = -\sqrt{3}$.
Three inflection points: $(0, 0)$, $(\sqrt{3}, \frac{\sqrt{3}}{4})$, $(-\sqrt{3}, -\frac{\sqrt{3}}{4})$.
Sketch: An odd function passing through the origin, rising to a local max of $1/2$ at $x = 1$, then decaying to $0$; symmetric behavior on the left with a local min of $-1/2$ at $x = -1$.
Problem 8. A farmer has 600 meters of fencing and wants to enclose a rectangular area along a straight river. No fencing is needed along the river. What dimensions maximize the enclosed area?
Show Solution
Let $x$ = length parallel to the river and $y$ = width perpendicular to the river.
Constraint: $x + 2y = 600$, so $x = 600 - 2y$.
Objective: $A = xy = (600 - 2y)y = 600y - 2y^2$.
Domain: $0 < y < 300$.
$A'(y) = 600 - 4y = 0 \implies y = 150$.
$A''(y) = -4 < 0$, so this is a maximum.
$x = 600 - 2(150) = 300$.
Answer: The area is maximized with dimensions $300 \text{ m} \times 150 \text{ m}$, giving $A = 45{,}000$ m$^2$.
Constraint: $x + 2y = 600$, so $x = 600 - 2y$.
Objective: $A = xy = (600 - 2y)y = 600y - 2y^2$.
Domain: $0 < y < 300$.
$A'(y) = 600 - 4y = 0 \implies y = 150$.
$A''(y) = -4 < 0$, so this is a maximum.
$x = 600 - 2(150) = 300$.
Answer: The area is maximized with dimensions $300 \text{ m} \times 150 \text{ m}$, giving $A = 45{,}000$ m$^2$.
Problem 9. Use the second derivative test to classify the critical points of $f(x) = \frac{x^2}{x - 2}$ (for $x \neq 2$).
Show Solution
First derivative:
$$f'(x) = \frac{2x(x-2) - x^2}{(x-2)^2} = \frac{x^2 - 4x}{(x-2)^2} = \frac{x(x-4)}{(x-2)^2}$$
$f'(x) = 0$ when $x = 0$ or $x = 4$.
Second derivative: (by quotient rule, after simplification) $$f''(x) = \frac{-8}{(x-2)^3}$$ Evaluate at critical points:
Second derivative: (by quotient rule, after simplification) $$f''(x) = \frac{-8}{(x-2)^3}$$ Evaluate at critical points:
- $f''(0) = \frac{-8}{(-2)^3} = \frac{-8}{-8} = 1 > 0$. Local minimum at $f(0) = 0$.
- $f''(4) = \frac{-8}{(2)^3} = \frac{-8}{8} = -1 < 0$. Local maximum at $f(4) = \frac{16}{2} = 8$.
Problem 10. Find the absolute maximum and absolute minimum of $f(x) = \sin x + \cos x$ on the interval $[0, 2\pi]$.
Show Solution
$f'(x) = \cos x - \sin x = 0 \implies \cos x = \sin x \implies \tan x = 1$.
On $[0, 2\pi]$: $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.
Evaluate $f$ at critical points and endpoints:
Absolute minimum: $-\sqrt{2}$ at $x = \frac{5\pi}{4}$.
On $[0, 2\pi]$: $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.
Evaluate $f$ at critical points and endpoints:
- $f(0) = 0 + 1 = 1$
- $f\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414$
- $f\!\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} \approx -1.414$
- $f(2\pi) = 0 + 1 = 1$
Absolute minimum: $-\sqrt{2}$ at $x = \frac{5\pi}{4}$.
Problem 11. Determine the intervals on which $h(x) = x\ln x$ ($x > 0$) is concave up and concave down.
Show Solution
$h'(x) = \ln x + 1$ (by the product rule).
$h''(x) = \frac{1}{x}$.
For $x > 0$: $h''(x) = \frac{1}{x} > 0$ for all $x$ in the domain.
Therefore $h$ is concave up on $(0, \infty)$. There are no inflection points and no intervals of concave down behavior on the domain.
$h''(x) = \frac{1}{x}$.
For $x > 0$: $h''(x) = \frac{1}{x} > 0$ for all $x$ in the domain.
Therefore $h$ is concave up on $(0, \infty)$. There are no inflection points and no intervals of concave down behavior on the domain.
Problem 12. A particle moves along the $x$-axis with position $x(t) = t^3 - 6t^2 + 9t + 2$ for $t \geq 0$. Find the times when the particle is at rest, the intervals when it moves to the right, and the times when it changes direction. Also determine when the particle's speed is increasing.
Show Solution
Velocity: $v(t) = x'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$.
At rest: $v(t) = 0$ when $t = 1$ and $t = 3$.
Sign of velocity:
Acceleration: $a(t) = v'(t) = 6t - 12 = 6(t - 2)$.
$a(t) = 0$ at $t = 2$; $a < 0$ for $t < 2$, $a > 0$ for $t > 2$.
Speed is increasing when velocity and acceleration have the same sign:
At rest: $v(t) = 0$ when $t = 1$ and $t = 3$.
Sign of velocity:
- $0 < t < 1$: $v > 0$ (moving right).
- $1 < t < 3$: $v < 0$ (moving left).
- $t > 3$: $v > 0$ (moving right).
Acceleration: $a(t) = v'(t) = 6t - 12 = 6(t - 2)$.
$a(t) = 0$ at $t = 2$; $a < 0$ for $t < 2$, $a > 0$ for $t > 2$.
Speed is increasing when velocity and acceleration have the same sign:
- $1 < t < 2$: $v < 0$ and $a < 0$. Same sign. Speed increasing.
- $t > 3$: $v > 0$ and $a > 0$. Same sign. Speed increasing.