# Sketching graphs of f, f', f''

> AP Calculus BC · Analytical Applications of Differentiation
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u5-sketching-graphs-of-f-f/

This subtopic connects the graphical behavior of a function, its first derivative, and its second derivative. You will learn to sketch or match any function given one of the three, and identify key features like extrema and inflection points for the AP exam.

**Prerequisites:** Computing first and second derivatives of common functions; Finding zeros of algebraic and transcendental functions; Formal definitions of critical points and inflection points

## Learning objectives

- Connect graphical behavior of f to f' and f''
- Sketch or match graphs of derivatives given any one function
- Identify key features (extrema, inflection points) from derivative graphs
- Avoid common exam traps related to derivative graph interpretation

## Relationship Between f and f'

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 $f'$:

1. When $f$ is increasing on an interval, $f'(x) > 0$, so $f'$ lies above the $x$-axis
2. When $f$ is decreasing on an interval, $f'(x) < 0$, so $f'$ lies below the $x$-axis
3. At local extrema of $f$, $f'(x) = 0$, so $f'$ crosses or touches the $x$-axis at that $x$-value
4. 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 $f(x)$.

1. Find critical points of $f$, which occur at $f'(x) = 0$. Compute the derivative:
2. $$f'(x) = 3x^2 - 3x - 18 = 3(x-3)(x+2)$$
3. 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:
4. For $x < -2$, $f'(x) > 0$, so $f$ is increasing here, meaning $f'(x)$ is positive (above the $x$-axis)
5. For $-2 < x < 3$, $f'(x) < 0$, so $f$ is decreasing here, meaning $f'(x)$ is negative (below the $x$-axis)
6. For $x > 3$, $f'(x) > 0$, so $f$ is increasing here, meaning $f'(x)$ is positive (above the $x$-axis)
7. 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.

> **tip**
>
> When matching $f$ from $f'$ on multiple choice, always eliminate options first by checking the sign of $f'$. Any option where $f$ is increasing when $f'$ is negative can be immediately crossed out, saving valuable exam time.

*Calculator:* forbidden

## Relationship Between f and f''

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:

1. When $f$ is concave up on an interval, $f''(x) > 0$, so $f''$ lies above the $x$-axis
2. When $f$ is concave down on an interval, $f''(x) < 0$, so $f''$ lies below the $x$-axis
3. At inflection points of $f$ (where concavity changes for continuous $f$), $f''(x) = 0$, so $f''$ crosses the $x$-axis at that $x$-value
4. 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?

1. 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 $(-\infty, 1)$
2. For $x > 1$, $f$ is concave up, so $f''(x) > 0$ for all $x > 1$, meaning $f''$ lies above the $x$-axis on $(1, \infty)$
3. $f$ has an inflection point at $x=1$, so $f''(1) = 0$, so $f''$ crosses the $x$-axis at $(1, 0)$
4. 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 $x=1$.

> **tip**
>
> On the AP exam, inflection points only occur where concavity of $f$ actually changes, not just where $f''(x) = 0$. Always confirm the sign of $f''$ changes on either side of $x=a$ before marking an inflection point at $a$.

*Calculator:* forbidden

## Matching and Sketching All Three Graphs

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:

1. 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
2. Split the $x$-axis into intervals separated by these key $x$-values
3. 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 $f$)
4. 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 $f''(x)$.

1. 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
2. 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 $x=2$
3. Next analyze $f''(x)$ from $f'(x)$: $f'$ is a downward-opening quadratic, so its derivative ($f''(x)$) is linear
4. 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 $(-\infty, 0.5)$
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 $x=0.5$
6. 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.

> **tip**
>
> When working from a given graph, write the sign of the given function above each interval first, then write the corresponding behavior of the target function below. This simple step eliminates 90% of common sign mix-ups.

*Calculator:* forbidden

## AP Style Concept Check

**Check your understanding**

Test your understanding with this multiple choice question:

1. The graph of $y = f''(x)$ crosses the $x$-axis at $x=-2$, $x=1$, and $x=3$. It is negative on $(-\infty, -2)$, positive on $(-2, 1)$, negative on $(1, 3)$, and positive on $(3, \infty)$. For the original function $f(x)$, how many inflection points does $f$ have?

   - 1
   - 2
   - 3
   - 0

   *Why:* Inflection points occur where $f''$ changes sign, which happens at every crossing of the $x$-axis here. All three points have a sign change, so $f$ has 3 inflection points.

**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.

1. Part (a): Find zeros of $f'(x)$: $2\sin x = 0$ at $x=0$, $x=\pi$, $x=2\pi$
2. $f'(x) > 0$ 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 $x=\pi$.
3. Part (b): Calculate $f''(x)$:
4. $$f''(x) = \frac{d}{dx}[2\sin x] = 2\cos x$$
5. 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.
6. Part (c): Find the antiderivative of $f'(x)$:
7. $$f(x) = -2\cos x + C$$
8. 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), $(2\pi, 2)$.

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Claiming $f$ has a local extremum at $x=a$ just because $f''(a) = 0$
  - Why it fails: Students confuse inflection point rules with extremum rules, mixing up what $f'$ and $f''$ tell you
  - Correct: Only conclude $f$ has a local extremum at $a$ if $f'$ changes sign at $a$; $f''(a) = 0$ only indicates a possible inflection point for $f$
- **Wrong:** Drawing $f$ as crossing the $x$-axis at the same $x$ where $f'$ crosses the $x$-axis
  - Why it fails: Students confuse x-intercepts of $f'$ with x-intercepts of $f$, matching all key points to the same $x$ across graphs regardless of meaning
  - Correct: Remember that x-intercepts of $f'$ correspond to local extrema (critical points) of $f$, not x-intercepts of $f$
- **Wrong:** Stating that $f$ is concave up where $f'$ is positive
  - Why it fails: Students mix up the meaning of first vs second derivative signs, conflating increasing/decreasing with concavity
  - Correct: Always associate $f'$ sign with $f$ increasing/decreasing, and $f''$ sign with $f$ concavity; write this association down on scratch paper for every problem
- **Wrong:** Drawing $f''$ as touching (not crossing) the $x$-axis at an inflection point of $f$ when concavity changes
  - Why it fails: Students memorize that crossing means sign change, but forget that a sign change of $f''$ is required for an inflection point
  - Correct: If concavity changes at $x=a$, $f''$ must change sign at $a$, so $f''$ crosses the $x$-axis at $a$; only a tangent that doesn't cross means no sign change, hence no inflection point
- **Wrong:** When given $f'$ and asked for $f$'s inflection points, looking for where $f'$ crosses the $x$-axis
  - Why it fails: Students confuse where $f'$ has zero value with where $f'$ has zero slope
  - Correct: Inflection points of $f$ occur at extrema of $f'$, where $f'$ changes slope, i.e., where $f'$ has a local maximum or minimum, not its x-intercepts
- **Wrong:** Assuming that if $f$ is increasing everywhere, then $f'$ is also increasing everywhere
  - Why it fails: Students assume increasing $f$ means positive $f'$, so positive $f'$ must be increasing
  - Correct: An increasing $f$ can have a decreasing positive slope (e.g., $f(x) = \ln x$ for $x>0$), so always check the slope of $f'$ to get behavior of $f''$

## Cheatsheet

| Relationship | Rule | Notes |
| --- | --- | --- |
| f, f' | If $f'(x) > 0$ on $(a,b)$, $f(x)$ increasing on $(a,b)$ | All differentiable $f$ |
| f, f' | If $f'(x) < 0$ on $(a,b)$, $f(x)$ decreasing on $(a,b)$ | All differentiable $f$ |
| f, f' | If $f'(c) = 0$ and $f'$ changes sign at $c$, $f$ has local extremum at $x=c$ | Zeros of $f'$ = critical points of $f$ |
| f, f'' | If $f''(x) > 0$ on $(a,b)$, $f(x)$ concave up on $(a,b)$ | Also means $f'$ is increasing on $(a,b)$ |
| f, f'' | If $f''(x) < 0$ on $(a,b)$, $f(x)$ concave down on $(a,b)$ | Also means $f'$ is decreasing on $(a,b)$ |
| f, f'' | If $f''(c) = 0$ and $f''$ changes sign at $c$, $f$ has inflection point at $x=c$ | $f''=0$ alone does not guarantee an inflection point |

## What's next

This subtopic is a core foundational skill for all analytical applications of differentiation on the AP Calculus BC exam. The relationships you master here transfer directly to optimization problems, full curve sketching of complex functions, and analyzing motion problems that appear regularly in both multiple choice and free response. Understanding how $f$, $f'$, and $f''$ connect also supports graphical interpretation of integration problems, which make up a large portion of the exam score. Building fluency with this skill will make more advanced applications of derivatives much easier to master.

- [Unit 5 Overview: Analytical Applications of Differentiation](https://www.owlsprep.com/study/ap-calculus-bc-u5-overview/)
- [Connecting f, f', f'' qualitatively](https://www.owlsprep.com/study/ap-calculus-bc-u5-connecting-f-f-f-qualitatively/)
- [Introduction to Optimization Problems](https://www.owlsprep.com/study/ap-calculus-bc-u5-introduction-to-optimization-problems/)

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