Chapter 5: Curve Sketching & Optimization
AP Calculus AB & BC | Unit 2: Applications of Derivatives
Learning Objectives
- Find and classify critical points using the First Derivative Test
- Apply the Second Derivative Test to identify local maxima and minima
- Analyze concavity and locate inflection points using $f''$
- Sketch curves using a systematic analysis of $f$, $f'$, and $f''$
- Set up and solve optimization problems to find absolute extrema
5.1 Critical Points and the First Derivative Test
A function's graph rises and falls based on the sign of its derivative. Where $f'$ changes sign, we find local extrema.
Definition: Critical Points
A number $c$ in the domain of $f$ is a critical point if $f'(c) = 0$ or $f'(c)$ does not exist.
Local extrema can only occur at critical points — but not every critical point is a local extremum.
First Derivative Test
Let $c$ be a critical point of a continuous function $f$:
- If $f'$ changes from positive to negative at $c$: local maximum at $c$.
- If $f'$ changes from negative to positive at $c$: local minimum at $c$.
- If $f'$ does not change sign at $c$: no local extremum (e.g., an inflection point).
Example 5.1 — Finding and Classifying Critical Points
Analyze $f(x) = x^3 - 3x^2 - 9x + 2$.
Step 1 — Find critical points: $f'(x) = 3x^2 - 6x - 9 = 3(x^2-2x-3) = 3(x-3)(x+1)$
Critical points: $x = -1$ and $x = 3$.
Step 2 — Sign chart for $f'$:
Conclusion: Local maximum at $x = -1$: $f(-1) = -1-3+9+2 = 7$. Local minimum at $x = 3$: $f(3) = 27-27-27+2 = -25$.
TRY IT
Find the critical points of $g(x) = x^4 - 4x^3$ and classify each using the First Derivative Test.
Show Answer
$g'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$
Critical points: $x = 0$ and $x = 3$.
Sign: $g' < 0$ on $(-\infty,0)$, $g'(0)=0$ (sign doesn't change), $g' < 0$ on $(0,3)$, $g'(3)=0$, $g' > 0$ on $(3,\infty)$.
At $x=0$: no extremum (sign unchanged). At $x=3$: local minimum, $g(3) = 81-108 = -27$.
$f(x) = x^3-3x^2-9x+2$ (blue) and $f'(x)$ (red). Where $f'$ crosses zero and changes sign, the original function has a local extremum.
Figure 5.1 — Critical Points: $f$ and $f'$ for $x^3-3x^2-9x+2$
5.2 Concavity and the Second Derivative Test
The second derivative $f''$ tells us about the rate of change of the slope — whether the curve bends upward (concave up) or downward (concave down).
Concavity and Inflection Points
- Concave up on $(a,b)$: $f''(x) > 0$ — graph curves like a bowl $\cup$. Tangent lines lie below the curve.
- Concave down on $(a,b)$: $f''(x) < 0$ — graph curves like a cap $\cap$. Tangent lines lie above the curve.
- Inflection point: a point where $f''$ changes sign (concavity changes). Find candidates where $f''(x) = 0$ or $f''(x)$ is undefined.
Second Derivative Test
If $f'(c) = 0$ (so $c$ is a critical point):
- If $f''(c) > 0$: local minimum at $c$ (concave up, so the critical point is a valley).
- If $f''(c) < 0$: local maximum at $c$ (concave down, so the critical point is a peak).
- If $f''(c) = 0$: test is inconclusive — use the First Derivative Test instead.
Example 5.2 — Second Derivative Test on $f(x) = x^3 - 3x^2 - 9x + 2$
From Example 5.1: critical points at $x = -1$ and $x = 3$.
$f''(x) = 6x - 6$
$f''(-1) = -6 - 6 = -12 < 0$ → local maximum at $x = -1$ ✓
$f''(3) = 18 - 6 = 12 > 0$ → local minimum at $x = 3$ ✓
Inflection point: $f''(x) = 0 \Rightarrow 6x - 6 = 0 \Rightarrow x = 1$. Check sign change: $f'' < 0$ for $x < 1$, $f'' > 0$ for $x > 1$. Inflection at $(1, f(1)) = (1, -9)$.
TRY IT
Find all inflection points of $h(x) = x^4 - 6x^2$.
Show Answer
$h'(x) = 4x^3 - 12x$
$h''(x) = 12x^2 - 12 = 12(x^2-1) = 12(x-1)(x+1)$
$h''=0$ at $x = \pm 1$. Check sign changes:
$h''(-2) = 36 > 0$, $h''(0) = -12 < 0$, $h''(2) = 36 > 0$
Sign changes at both $x = -1$ and $x = 1$. Inflection points at $(-1, -5)$ and $(1, -5)$.
★
AP Exam Tip: A common error is assuming that $f''(c)=0$ automatically means an inflection point. You must verify that $f''$ changes sign at $c$. For example, $f(x)=x^4$ has $f''(0)=0$ but no inflection point at $x=0$ since $f''$ doesn't change sign.
$f(x)$, $f'(x)$, and $f''(x)$ together. Where $f''$ changes sign, $f$ has an inflection point. Where $f''=0$ and $f'=0$, the Second Derivative Test applies.
Figure 5.2 — Concavity: $f$, $f'$, and $f''$ for $x^3-3x^2-9x+2$
5.3 Optimization Problems
Optimization problems ask for the maximum or minimum value of a quantity subject to a constraint. The key is to express the objective function (quantity to optimize) in terms of a single variable, then use calculus.
Optimization Procedure
- Identify the quantity to maximize/minimize (objective function $Q$).
- Identify the constraint — an equation relating the variables.
- Reduce to one variable using the constraint.
- Find critical points of $Q$ on the relevant domain.
- Test critical points and endpoints to find the absolute extremum.
Example 5.3 — Maximizing Area of a Fenced Rectangle
A farmer has 200 m of fencing to enclose a rectangular field against a barn wall (one side needs no fence). Maximize the area.
Variables: Width $x$ (two sides), length $y$ (one side). Constraint: $2x + y = 200 \Rightarrow y = 200 - 2x$.
Objective: $A = xy = x(200-2x) = 200x - 2x^2$, for $0 < x < 100$.
Optimize: $\dfrac{dA}{dx} = 200 - 4x = 0 \Rightarrow x = 50$
$\dfrac{d^2A}{dx^2} = -4 < 0$ → maximum ✓
$y = 200 - 100 = 100$ m. Maximum area $= 50 \times 100 = 5000$ m².
TRY IT
Find two positive numbers whose sum is 20 and whose product is as large as possible.
Show Answer
Let the numbers be $x$ and $20-x$. Maximize $P = x(20-x) = 20x - x^2$.
$P'(x) = 20 - 2x = 0 \Rightarrow x = 10$
$P''(10) = -2 < 0$ → maximum. Both numbers are $10$; product $= 100$.
Example 5.4 — Minimizing Material for a Can
A cylindrical can must hold $500\pi$ cm³. Find the radius and height that minimize total surface area.
Constraint: $V = \pi r^2 h = 500\pi \Rightarrow h = \dfrac{500}{r^2}$
Objective: $S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \dfrac{1000\pi}{r}$
Optimize: $\dfrac{dS}{dr} = 4\pi r - \dfrac{1000\pi}{r^2} = 0 \Rightarrow 4r^3 = 1000 \Rightarrow r^3 = 250 \Rightarrow r = \sqrt[3]{250} \approx 6.30$ cm
$h = \dfrac{500}{r^2} = \dfrac{500}{250^{2/3}} = 2\sqrt[3]{250} \approx 12.60$ cm
Note: $h = 2r$ — the optimal can has height equal to its diameter!
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AP Exam Tip: For optimization on a closed interval, always check the endpoints in addition to the critical points — the absolute maximum or minimum may occur at an endpoint. Use the Closed Interval Method: evaluate $f$ at all critical points and both endpoints, then compare.
Fenced rectangle optimization: drag the slider to change $x$ (width). The area curve peaks at the optimal $x = 50$.
Figure 5.3 — Optimization: Maximizing Area with Fixed Perimeter
Practice Problems
1
Find all critical points of $f(x) = 2x^3 - 9x^2 + 12x - 4$ and classify each as a local max, min, or neither.
Show Solution
$f'(x) = 6x^2-18x+12 = 6(x-1)(x-2)$
Critical points: $x=1$ and $x=2$.
$f''(x) = 12x-18$: $f''(1)=-6<0$ → local max; $f''(2)=6>0$ → local min.
Local max at $(1, 1)$; local min at $(2, 0)$.
2
Find the intervals on which $f(x) = x^4 - 8x^2$ is concave up and concave down, and find all inflection points.
Show Solution
$f''(x) = 12x^2 - 16 = 4(3x^2-4)$
$f''=0$: $x = \pm\dfrac{2}{\sqrt{3}} = \pm\dfrac{2\sqrt{3}}{3}$
Concave up: $\left(-\infty, -\tfrac{2\sqrt{3}}{3}\right)$ and $\left(\tfrac{2\sqrt{3}}{3}, \infty\right)$.
Concave down: $\left(-\tfrac{2\sqrt{3}}{3}, \tfrac{2\sqrt{3}}{3}\right)$.
Inflection points at $x = \pm\dfrac{2\sqrt{3}}{3}$.
3
Find the absolute maximum and minimum values of $g(x) = x^3 - 3x$ on $[-2, 3]$.
Show Solution
$g'(x) = 3x^2-3 = 3(x-1)(x+1)$. Critical points in $[-2,3]$: $x=\pm 1$.
Evaluate: $g(-2)=-2$, $g(-1)=2$, $g(1)=-2$, $g(3)=18$.
Absolute max: 18 at $x=3$; Absolute min: $-2$ at $x=-2$ and $x=1$.
4
A rectangular box (no lid) has a square base. If the total surface area is 300 cm², find the dimensions that maximize the volume.
Show Solution
Let base side $= x$, height $= h$. Surface: $x^2 + 4xh = 300 \Rightarrow h = \dfrac{300-x^2}{4x}$
Volume: $V = x^2 h = \dfrac{x(300-x^2)}{4} = 75x - \dfrac{x^3}{4}$
$V'(x) = 75 - \dfrac{3x^2}{4} = 0 \Rightarrow x^2 = 100 \Rightarrow x = 10$
$h = \dfrac{300-100}{40} = 5$. Box: 10 cm × 10 cm × 5 cm.
5
Sketch the curve $f(x) = x^3 - 6x^2 + 9x$. Identify all local extrema and inflection points.
Show Solution
$f'(x) = 3x^2-12x+9 = 3(x-1)(x-3)$. Critical: $x=1,3$.
$f(1)=4$ (local max), $f(3)=0$ (local min).
$f''(x) = 6x-12 = 0 \Rightarrow x=2$. $f(2)=2$. Inflection at $(2,2)$.
Concave down for $x<2$, concave up for $x>2$.
6
Find the point on the curve $y = \sqrt{x}$ closest to the point $(3, 0)$.
Show Solution
Minimize $D^2 = (x-3)^2 + y^2 = (x-3)^2 + x$ (easier than minimizing $D$).
Let $Q(x) = (x-3)^2 + x$. $Q'(x) = 2(x-3)+1 = 2x-5 = 0 \Rightarrow x = 5/2$.
$y = \sqrt{5/2} = \dfrac{\sqrt{10}}{2}$. Closest point: $\left(\dfrac{5}{2}, \dfrac{\sqrt{10}}{2}\right)$.
7
The second derivative of $f$ is $f''(x) = x^2(x-2)(x+3)$. Find the intervals of concavity and inflection points of $f$.
Show Solution
Zeros of $f''$: $x=-3, 0, 2$. Analyze sign:
$(-\infty,-3)$: test $x=-4$: $(-4)^2(-6)(-1)>0$ → concave up
$(-3,0)$: test $x=-1$: $(1)(-3)(-2)>0$... $(-1)^2(-3)(-2)=6>0$... wait, sign of $(x-2)(x+3)$ at $x=-1$: $(-3)(2)=-6<0$. So $f''<0$ → concave down
$(0,2)$: $(x-2)(x+3)<0$ for $x\in(0,2)$: same sign → concave down
$(2,\infty)$: both factors positive → concave up
Inflection points at $x=-3$ (sign change) and $x=2$ (sign change). At $x=0$: no sign change → no inflection.
8
(AP Style) A particle moves along a line with position $s(t) = t^3 - 6t^2 + 9t$ for $t \geq 0$. Find when the particle is moving right, when it's moving left, and find its total distance traveled on $[0, 4]$.
Show Solution
$v(t) = s'(t) = 3t^2-12t+9 = 3(t-1)(t-3)$
Moving right ($v>0$): $t\in[0,1)$ and $t\in(3,4]$.
Moving left ($v<0$): $t\in(1,3)$.
Total distance = $|s(1)-s(0)| + |s(3)-s(1)| + |s(4)-s(3)|$
$s(0)=0$, $s(1)=4$, $s(3)=0$, $s(4)=4$
Total $= |4-0|+|0-4|+|4-0| = 4+4+4 = 12$ units.
📋 Chapter Summary
Core Concepts
First Derivative Test
At a critical point $c$: if $f'$ changes $+\to-$, local max; if $f'$ changes $-\to+$, local min; if no sign change, neither.
Second Derivative Test
At critical point $c$ where $f'(c)=0$: if $f''(c)>0$, local min; if $f''(c)<0$, local max; if $f''(c)=0$, inconclusive.
Concavity
$f''>0$ on an interval $\Rightarrow$ concave up (cup shape). $f''<0$ $\Rightarrow$ concave down (cap shape).
Closed Interval Method (Optimization)
On $[a,b]$, find all critical points and evaluate $f$ at critical points and endpoints. The largest value is the absolute max, smallest is the absolute min.
Key Definitions
Critical Point
A point where $f'(c) = 0$ or $f'(c)$ does not exist — candidates for local extrema.
Inflection Point
A point where $f''$ changes sign — concavity switches from up to down or vice versa.
Absolute Extrema
The global maximum or minimum value of $f$ over its entire domain or a specified interval.
Increasing/Decreasing
$f'>0$ on an interval $\Rightarrow$ $f$ is increasing. $f'<0$ $\Rightarrow$ $f$ is decreasing.
Curve Sketching Checklist
- Domain — find where $f$ is defined
- Intercepts — set $f(x)=0$ and find $f(0)$
- Asymptotes — check for vertical ($f\to\infty$) and horizontal ($\lim_{x\to\pm\infty}$) asymptotes
- Increasing/Decreasing — find $f'$, critical points, sign chart
- Local Extrema — first or second derivative test at critical points
- Concavity & Inflection — find $f''$, sign chart, inflection points
📘 Key Terms
Critical Point
An input $c$ in the domain where $f'(c) = 0$ or $f'(c)$ is undefined — candidates for local extrema.
Local Maximum
A value $f(c)$ that is greater than all nearby function values. Found via first or second derivative test.
Local Minimum
A value $f(c)$ that is less than all nearby function values.
Inflection Point
A point on the graph where concavity changes from up to down (or vice versa) — where $f''$ changes sign.
Concave Up / Down
Concave up: $f''>0$, graph curves like a cup. Concave down: $f''<0$, graph curves like a cap.
Optimization
Using calculus to find the absolute maximum or minimum of a function subject to constraints.