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

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

This module covers the graphical relationships between a function $f$, its first derivative $f'$, and second derivative $f''$, including how to match graphs, identify key features, and sketch one graph from information about another, a heavily tested AP Calculus AB topic.

**Prerequisites:** Computing first and second derivatives of common functions; Classifying critical points and inflection points via derivative tests; Identifying core graph features (intercepts, slopes)

## Learning objectives

- Relate key graphical features of f to f' and f''
- Identify extrema and inflection points across derivative graphs
- Sketch f given a graph of f' and initial condition
- Solve common AP exam matching and FRQ problems for this topic

## Relating Key Features of f and f'

The first derivative $f'$ gives the slope of the tangent line to $f$ at any point $x$, so every key behavior of $f$ translates directly to a graphical feature of $f'$, and vice versa. This topic tests your conceptual understanding of what derivatives measure, rather than just computation.

- When $f$ is increasing on an interval, all tangent slopes are positive, so $f'(x) > 0$ and the graph of $f'$ lies above the $x$-axis.
- When $f$ is decreasing on an interval, all tangent slopes are negative, so $f'(x) < 0$ and the graph of $f'$ lies below the $x$-axis.
- Local maxima/minima of $f$ (for differentiable $f$) occur where $f'$ crosses the $x$-axis: a local maximum of $f$ means $f'$ changes from positive to negative, and a local minimum means $f'$ changes from negative to positive.
- For polynomials, differentiation reduces degree by 1: a degree $n$ $f$ has a degree $n-1$ $f'$, a quick check for multiple-choice matching.

**Worked example:** The graph of a differentiable cubic function $f(x) = x^3 - 12x$ has critical points at $x=-2$ and $x=2$. Use the graph of $f$ to sketch the key features of $f'$.

1. First, identify intervals of increase/decrease for $f$: For $x < -2$, $f$ is increasing, so $f'(x) > 0$, meaning $f'$ lies above the $x$-axis on $(-\infty, -2)$.
2. For $-2 < x < 2$, $f$ decreases across this interval, so $f'(x) < 0$, and $f'$ lies below the $x$-axis.
3. For $x > 2$, $f$ increases on $(2, \infty)$, so $f'(x) > 0$, and $f'$ lies above the $x$-axis.
4. $f'$ has $x$-intercepts at $x=-2$ and $x=2$, matching the critical points of $f$. Since $f$ is degree 3, $f'$ is an upward-opening degree 2 parabola, which matches our sign analysis.

> **tip**
>
> On multiple-choice matching problems, check the sign of $f'$ (based on whether $f$ is increasing or decreasing) first—this eliminates 2-3 wrong answers immediately, saving time on test day.

## Relating Features of f and f' to f''

The second derivative $f''$ is the derivative of $f'$, so the same $f$-$f'$ relationship applies between $f'$ and $f''$. In addition, $f''$ encodes the concavity of the original function $f$, a key graphical property.

- When $f$ is concave up, it bends upward, the slope of $f$ is increasing, so $f''(x) > 0$.
- When $f$ is concave down, it bends downward, the slope of $f$ is decreasing, so $f''(x) < 0$.
- An inflection point occurs where $f$ changes concavity, which means $f''$ changes sign at that point. For twice-differentiable $f$, this also means $f'$ has a local extremum at that $x$-coordinate.

**Worked example:** Given $f(x) = x^4 - 6x^2$, find the $x$-coordinates of inflection points of $f$ by analyzing the graph of $f'$, then confirm with $f''$.

1. First compute $f' = 4x^3 - 12x$, a cubic function with $x$-intercepts at $x=-\sqrt{3}, 0, \sqrt{3}$. Inflection points of $f$ correspond to local extrema of $f'$, so we analyze where $f'$ increases and decreases.
2. Test intervals: $f'$ increases when $x < -1$, decreases when $-1 < x < 1$, and increases again when $x > 1$. So $f'$ has a local maximum at $x=-1$ and a local minimum at $x=1$.
3. Because $f'$ changes from increasing to decreasing at $x=-1$, and from decreasing to increasing at $x=1$, $f''$ changes sign at both $x=\pm 1$, meaning $f$ changes concavity at these points.
4. Confirm directly: $f''(x) = 12x^2 - 12 = 12(x^2 - 1)$, which crosses the $x$-axis at $x=\pm 1$ and changes sign at both points, matching our analysis from $f'$.

> **tip**
>
> If asked for inflection points of $f$ from a graph of $f'$, always look for the local maxima/minima of $f'$, not $x$-intercepts of $f'$—this is the most common mistake on this type of question.

## Sketching f From a Given Graph of f'

A very common AP Calculus AB free-response question gives you the graph of $f'$ (often piecewise linear for easy analysis) and an initial condition $f(a) = k$, then asks you to sketch $f$ or identify its key features. Follow this step-by-step process:

1. Split the $x$-axis into intervals separated by $x$-intercepts of $f'$ (these are critical points of $f$).
2. Find the sign of $f'$ on each interval to classify critical points as local maximum, minimum, or neither.
3. Split intervals further by local extrema of $f'$ (these are inflection points of $f$), then find concavity of $f$ on each interval.
4. Calculate $y$-values of key points of $f$ using the Fundamental Theorem of Calculus: $f(b) = f(a) + \int_a^b f'(x) dx$, which simplifies to adding areas of geometric shapes when $f'$ is piecewise linear.

**Worked example:** The graph of $f'$ is a straight line passing through $(0, 4)$ and $(2, 0)$. Given $f(0) = 1$, sketch $f$, identifying all key features.

1. Find the equation of $f'$: slope is $\frac{0-4}{2-0} = -2$, so $f'(x) = -2x + 4$. For $x < 2$, $f'(x) > 0$, so $f$ is increasing; for $x > 2$, $f'(x) < 0$, so $f$ is decreasing. The critical point is at $x=2$.
2. Classify the critical point: $f'$ changes from positive to negative at $x=2$, so $f$ has a local maximum at $x=2$. Calculate the $y$-value:
3. $$f(2) = f(0) + \text{area under } f' \text{ from } 0 \text{ to } 2 = 1 + \frac{(2)(4)}{2} = 5$$
4. So the local maximum is at $(2, 5)$. Find concavity: $f''(x)$ is the constant slope of $f' = -2$, which is always negative, so $f$ is always concave down, with no inflection points.
5. Final sketch: Start at $(0, 1)$, increase with decreasing slope to $(2, 5)$, then decrease with increasingly negative slope, remaining concave down everywhere.

> **tip**
>
> When $f'$ is piecewise linear, always use geometric area to find $y$-values of $f$ instead of integrating—this is faster and less error-prone on exam day.

## AP-Style Concept Check

**Check your understanding**

Test your understanding of core relationships with this AP-style multiple-choice question:

1. The graph of a twice-differentiable function $f$ has a local minimum at $x=3$ and changes concavity from down to up at $x=-1$. Which of the following must be true?

   - A) $f'(3) = 0$ and $f''(-1) > 0$
   - B) $f'(3) = 0$, $f''(-1) = 0$, and $f''$ changes from negative to positive at $x=-1$
   - C) $f'(3) > 0$, $f''(-1) = 0$, and $f''$ changes from negative to positive at $x=-1$
   - D) $f'(3) = 0$, $f''(-1) = 0$, and $f''$ changes from positive to negative at $x=-1$

   *Answer:* B) $f'(3) = 0$, $f''(-1) = 0$, and $f''$ changes from negative to positive at $x=-1$

   *Why:* Any local extremum of a differentiable function has $f'(x)=0$, eliminating option C. A concavity change from down to up means $f''$ changes from negative to positive at the inflection point, so B is the correct answer.

## Common pitfalls

- **Wrong:** Looking for x-intercepts on f' to find inflection points of f
  - Why it fails: Students confuse critical points of f (which are x-intercepts of f') with inflection points of f (which are x-intercepts of f'' or local extrema of f').
  - Correct: Memorize: x-intercepts of f' = critical points of f; local extrema of f' = inflection points of f.
- **Wrong:** Assuming f must have an inflection point at x=c just because f''(c) = 0
  - Why it fails: Inflection points require a sign change of f'', not just a zero value. For example, $f(x) = x^4$ has $f''(0) = 0$ but no sign change, so no inflection point.
  - Correct: Always check that f'' changes sign around x=c after finding f''(c)=0 before confirming an inflection point.
- **Wrong:** Swapping the sign of f' when working from f: marking f' negative when f is increasing
  - Why it fails: Time pressure on the exam leads to flipped relationships when working backwards.
  - Correct: Write the base rule 'f increasing → f' positive, f decreasing → f' negative' at the top of your scratch paper before starting matching problems.
- **Wrong:** Drawing f with an inflection point at the x-intercept of f' when sketching from f'
  - Why it fails: This comes from confusing critical points and inflection points under time pressure.
  - Correct: When sketching f from f', mark inflection points of f at the local maxima/minima of f', not at the x-intercepts of f'.
- **Wrong:** Claiming a cubic f must have a cubic f'
  - Why it fails: Students forget that differentiation lowers the degree of a polynomial by 1.
  - Correct: For polynomial matching problems, first check the degree of f to eliminate any options for f' that have the wrong degree.

## Cheatsheet

| Category | Relationship | Notes |
| --- | --- | --- |
| f monotonicity and f' | $f$ increasing on $I \iff f'(x) > 0$ for all $x \in I$; $f$ decreasing on $I \iff f'(x) < 0$ for all $x \in I$ | Applies to all differentiable f; constant f $\iff f'(x) = 0$ |
| Local extrema of f | Local extremum at $x=c \implies f'(c) = 0$ for differentiable f. Local max: f' changes + to -; Local min: f' changes - to + | An x-intercept of f' does not guarantee an extremum; a sign change is required |
| f concavity and f'' | $f$ concave up on $I \iff f''(x) > 0$ for all $x \in I$; $f$ concave down on $I \iff f''(x) < 0$ for all $x \in I$ | Concave up: tangents lie below f; concave down: tangents lie above f |
| Inflection points of f | Inflection point at $x=c \iff$ f changes concavity at $c \iff$ f'' changes sign at $c \iff$ f' has local extremum at $c$ | $f''(c) = 0$ alone is not sufficient; a sign change is required |
| y-values of f from f' | $f(b) = f(a) + \int_a^b f'(x) dx$ | For piecewise linear f', use geometric area of triangles/rectangles to compute the integral |
| Polynomial degree relationship | If f is degree $n$, f' is degree $n-1$, f'' is degree $n-2$ | Useful for eliminating wrong options in MCQ matching problems |
| Sign change summary | Local extremum of f: f' changes sign; Inflection point of f: f'' changes sign | Always check for sign changes, not just zero values |

## What's next

This topic is the conceptual foundation for all further work with derivatives and graphical analysis in AP Calculus AB. Next, you will apply these relationships to full curve sketching of single-variable functions and optimization problems, where you find the maximum or minimum of a real-world function by classifying its critical points. Without mastering the correspondence between $f$, $f'$, and $f''$ features, you cannot correctly classify extrema or interpret results in optimization problems, which are a common high-weight FRQ topic. This skill also transfers directly to work with integration, where you will repeatedly sketch antiderivative graphs from derivative graphs, using the same area calculation and feature mapping you practiced here.

- [Connecting f, f', f'' qualitatively](https://www.owlsprep.com/study/ap-calculus-ab-u5-connecting-f-f-f-qualitatively/)
- [Introduction to Optimization Problems](https://www.owlsprep.com/study/ap-calculus-ab-u5-introduction-to-optimization-problems/)
- [Solving Optimization Problems](https://www.owlsprep.com/study/ap-calculus-ab-u5-solving-optimization-problems/)

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