Calculus BC · Analytical Applications of Differentiation · 14 min read · Updated 2026-05-11

Sketching graphs of f, f', f'' — AP Calculus BC

AP Calculus BC · Analytical Applications of Differentiation · 14 min read

1. Relationship Between f and f' ★★☆☆☆ ⏱ 4 min

Every point on the graph of $f'(x)$ equals the slope of the tangent line to $f(x)$ at the same $x$-value. This core relationship gives consistent rules connecting the behavior of $f$ to $f'$:

When $f$ is increasing on an interval, $f'(x) > 0$, so $f'$ lies above the $x$-axis
When $f$ is decreasing on an interval, $f'(x) < 0$, so $f'$ lies below the $x$-axis
At local extrema of $f$, $f'(x) = 0$, so $f'$ crosses or touches the $x$-axis at that $x$-value
If $f$ is linear, its slope is constant, so $f'$ forms a horizontal line

📐 Worked Example

Given $f(x) = x^3 - \frac{3}{2}x^2 - 18x + 10$, sketch the graph of $f'(x)$ by analyzing the behavior of $f(x)$.

Find critical points of $f$, which occur at $f'(x) = 0$. Compute the derivative:
$f'(x) = 3x^2 - 3x - 18 = 3(x-3)(x+2)$
Critical points of $f$ are at $x=-2$ and $x=3$, so $f'(x)$ crosses the $x$-axis at these $x$-values. Test the sign of $f'$ on each interval:
For $x < -2$, $f'(x) > 0$, so $f$ is increasing here, meaning $f'(x)$ is positive (above the $x$-axis)
For $-2 < x < 3$, $f'(x) < 0$, so $f$ is decreasing here, meaning $f'(x)$ is negative (below the $x$-axis)
For $x > 3$, $f'(x) > 0$, so $f$ is increasing here, meaning $f'(x)$ is positive (above the $x$-axis)
Since $f$ is cubic, $f'$ is quadratic with a positive leading coefficient. The final sketch is an upward-opening parabola crossing the $x$-axis at $x=-2$ and $x=3$, matching our sign intervals.

2. Relationship Between f and f'' ★★☆☆☆ ⏱ 4 min

Just as $f'$ describes the slope of $f$, $f''$ describes the slope of $f'$, which corresponds directly to the concavity of $f$. Concavity describes the direction a curve bends: concave up curves bend upward like a cup, and concave down curves bend downward like a cap. The core rules are:

When $f$ is concave up on an interval, $f''(x) > 0$, so $f''$ lies above the $x$-axis
When $f$ is concave down on an interval, $f''(x) < 0$, so $f''$ lies below the $x$-axis
At inflection points of $f$ (where concavity changes for continuous $f$), $f''(x) = 0$, so $f''$ crosses the $x$-axis at that $x$-value
If $f$ has constant concavity, $f''$ is constant, so $f''$ forms a horizontal line

📐 Worked Example

The graph of $f(x)$ has an inflection point at $x=1$, is concave down on $(-\infty, 1)$, and concave up on $(1, \infty)$. If $f(x)$ is a cubic polynomial, what does the graph of $f''(x)$ look like?

Concavity of $f$ directly translates to the sign of $f''$. Since $f$ is concave down for $x < 1$, $f''(x) < 0$ for all $x < 1$, so $f''$ lies below the $x$-axis on $(-\infty, 1)$
For $x > 1$, $f$ is concave up, so $f''(x) > 0$ for all $x > 1$, meaning $f''$ lies above the $x$-axis on $(1, \infty)$
$f$ has an inflection point at $x=1$, so $f''(1) = 0$, so $f''$ crosses the $x$-axis at $(1, 0)$
Since $f$ is cubic, its second derivative is linear. The sign of $f''$ goes from negative left of 1 to positive right of 1, so the slope of the line is positive. The final graph is a straight line with positive slope crossing the $x$-axis at $x=1$.

3. Matching and Sketching All Three Graphs ★★★☆☆ ⏱ 5 min

Most AP exam problems for this topic give you the graph of one of $f$, $f'$, or $f''$ and ask you to identify or sketch one of the other two. Follow this standardized step-by-step process:

Label all key $x$-values on the given graph: any $x$ where the graph crosses the $x$-axis, has a local extremum, or changes direction
Split the $x$-axis into intervals separated by these key $x$-values
For each interval, find the sign of the given graph, which tells you whether the target function is increasing/decreasing (if given $f'$) or concave up/down (if given $f$)
Connect key points to get the full graph, checking that the slope of the current graph matches the value of the derivative graph

📐 Worked Example

Given $f'(x)$ is a downward-opening quadratic crossing the $x$-axis at $x=-1$ and $x=2$, with $f'(x) > 0$ between $x=-1$ and $x=2$, and $f'(x) < 0$ for $x < -1$ and $x > 2$. Identify the key features of $f(x)$ and $f''(x)$.

First analyze $f(x)$ from $f'(x)$: Zeros of $f'$ are at $x=-1$ and $x=2$, so $f$ has critical points at these $x$-values
For $x < -1$, $f'(x) < 0$, so $f$ is decreasing. For $-1 < x < 2$, $f'(x) > 0$, so $f$ is increasing. For $x > 2$, $f'(x) < 0$, so $f$ is decreasing. Thus $f$ has a local minimum at $x=-1$ and a local maximum at $x=2$
Next analyze $f''(x)$ from $f'(x)$: $f'$ is a downward-opening quadratic, so its derivative ($f''(x)$) is linear
The vertex (maximum) of $f'$ is at $x=0.5$, the midpoint of $-1$ and $2$. For $x < 0.5$, $f'$ is increasing, so $f''(x) > 0$, meaning $f$ is concave up on $(-\infty, 0.5)$
For $x > 0.5$, $f'$ is decreasing, so $f''(x) < 0$, meaning $f$ is concave down on $(0.5, \infty)$. $f'$ has a maximum at $x=0.5$, so $f''(0.5) = 0$, which means $f$ has an inflection point at $x=0.5$
Final summary: $f''$ is a straight line with negative slope crossing the $x$-axis at $x=0.5$, and $f$ has key points at $x=-1$ (local min), $x=0.5$ (inflection point), and $x=2$ (local max) with the correct behavior in each interval.

4. AP Style Concept Check ★★★☆☆ ⏱ 5 min

📐 Worked Example

Let $f'(x) = 2\sin x$ on $0 \leq x \leq 2\pi$. (a) Find intervals where $f(x)$ is increasing/decreasing and all local extrema. (b) Find intervals of concavity and all inflection points. (c) Given $f(0) = 2$, find coordinates of all key points for sketching.

Part (a): Find zeros of $f'(x)$: $2\sin x = 0$ at $x=0$, $x=\pi$, $x=2\pi$
$f'(x) > 0$ on $(0, \pi)$, so $f$ is increasing on $(0, \pi)$. $f'(x) < 0$ on $(\pi, 2\pi)$, so $f$ is decreasing on $(\pi, 2\pi)$. By the First Derivative Test, $f$ has a local maximum at $x=\pi$.
Part (b): Calculate $f''(x)$:
$f''(x) = \frac{d}{dx}[2\sin x] = 2\cos x$
Zeros of $f''$ are at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$. $f''(x) > 0$ on $\left(0, \frac{\pi}{2}\right) \cup \left(\frac{3\pi}{2}, 2\pi\right)$, so $f$ is concave up here. $f''(x) < 0$ on $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$, so $f$ is concave down here. Both points have a sign change, so they are inflection points.
Part (c): Find the antiderivative of $f'(x)$:
$f(x) = -2\cos x + C$
Substitute $f(0) = 2$: $-2(1) + C = 2 \implies C=4$, so $f(x) = -2\cos x + 4$. Key points are $(0, 2)$, $\left(\frac{\pi}{2}, 4\right)$ (inflection), $(\pi, 6)$ (local max), $\left(\frac{3\pi}{2}, 4\right)$ (inflection), $(2\pi, 2)$.

Common Pitfalls

Why: Students confuse inflection point rules with extremum rules, mixing up what $f'$ and $f''$ tell you

Why: Students confuse x-intercepts of $f'$ with x-intercepts of $f$, matching all key points to the same $x$ across graphs regardless of meaning

Why: Students mix up the meaning of first vs second derivative signs, conflating increasing/decreasing with concavity

Why: Students memorize that crossing means sign change, but forget that a sign change of $f''$ is required for an inflection point

Why: Students confuse where $f'$ has zero value with where $f'$ has zero slope

Why: Students assume increasing $f$ means positive $f'$, so positive $f'$ must be increasing

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Sketching graphs of f, f', f'' — AP Calculus BC

1. Relationship Between f and f' ★★☆☆☆ ⏱ 4 min

2. Relationship Between f and f'' ★★☆☆☆ ⏱ 4 min

3. Matching and Sketching All Three Graphs ★★★☆☆ ⏱ 5 min

4. AP Style Concept Check ★★★☆☆ ⏱ 5 min

Common Pitfalls

Quick Reference Cheatsheet

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