Sketching graphs of f, f', f'' — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Relationships between the graphs of a function f, its first derivative f', and second derivative f''; identifying key features (extrema, inflection points, monotonicity, concavity) across graphs; constructing one graph from given information about another.

You should already know: How to compute first and second derivatives of common functions. How to classify critical points and inflection points via derivative tests. How to identify core graph features (intercepts, slopes) from function definitions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Sketching graphs of f, f', f''?

This topic is part of Unit 5: Analytical Applications of Differentiation in the AP Calculus AB Course and Exam Description (CED), and accounts for approximately 8-10% of total exam score weight, appearing in both multiple-choice (MCQ) and free-response (FRQ) sections. Unlike basic curve sketching of a single function from its algebraic equation, this topic focuses on the graphical correspondence between a function and its first two derivatives. You will be asked to match graphs, construct one graph from information about another, and interpret features across graphs, testing your conceptual understanding of what derivatives measure rather than just your ability to compute them. Common exam questions include matching f to f' or f'', identifying inflection points of f from a graph of f', and sketching f given a graph of f' and an initial condition. This is one of the most frequently tested conceptual topics on the AP exam, as it probes mastery of core derivative properties.

2. Relating Features of f to Features of 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. The core relationships are intuitive: when f increases across an interval, all tangent slopes are positive, so f'(x) > 0 and the graph of f' lies entirely above the x-axis on that interval. When f decreases, tangent slopes are negative, so f'(x) < 0 and f' lies below the x-axis. For differentiable f, any local maximum or minimum of f occurs where f crosses from increasing to decreasing or vice versa, so f' crosses the x-axis at the x-coordinate of f's extremum: if f has a local maximum at x=c, f' changes from positive to negative, so it crosses the x-axis from above to below. If f has a local minimum at x=c, f' changes from negative to positive, crossing from below to above. Conversely, x-intercepts of f' correspond to horizontal tangents and potential extrema on f. For polynomial functions, differentiation reduces the degree of the function by 1, so a degree n f becomes a degree n-1 f', which is a quick check for matching problems.

Worked Example

Problem: The graph of a differentiable cubic function f(x) = x³ - 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, test x=-3: f(-3) = (-27) - 12(-3) = 9, f(-2) = (-8) -12(-2) = 16, so f increases on (-∞, -2). This means f'(x) > 0 on (-∞, -2), so f' lies above the x-axis here.
2. For -2 < x < 2, test x=0: f(-2)=16, f(0)=0, f(2)=8-24=-16, so f decreases across (-2, 2). This means f'(x) < 0 here, so f' lies below the x-axis.
3. For x > 2, test x=3: f(2) = -16, f(3)=27-36=-9, so f increases on (2, ∞). This means f'(x) > 0 here, so 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 degree 2, an upward-opening parabola, which matches our sign analysis.

Exam 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.

3. Relating Features of f and f' to Features of f''

The second derivative f'' is the derivative of f', so the same relationship between f and f' applies between f' and f''. In addition, f'' encodes the concavity of f, a key graphical property of the original function. Concavity describes the direction the curve bends: when f is concave up, it bends upward, and the slope of f is increasing, which means f''(x) > 0 (since f'' is the rate of change of f'). When f is concave down, it bends downward, the slope of f is decreasing, so f''(x) < 0. An inflection point is where f changes concavity, which means f'' must change sign at that point. For twice-differentiable f, this means f'' crosses the x-axis at the inflection point of f, which also means f' has a local extremum (maximum or minimum) at that x-coordinate, since f' changes from increasing to decreasing or vice versa.

Worked Example

Problem: Given f(x) = x⁴ - 6x², find the x-coordinates of inflection points of f by analyzing the graph of f', then confirm with f''.

1. First compute f' to get its key features: f'(x) = 4x³ - 12x, a cubic function with x-intercepts at x=-√3, 0, √3. Inflection points of f correspond to local extrema of f', so we analyze where f' increases and decreases.
2. Test intervals: For x < -1, take x=-2: f''(-2) = 12(-2)² - 12 = 36 > 0, so f' increases here. For -1 < x < 1, take x=0: f''(0) = -12 < 0, so f' decreases here. For x > 1, take x=2: f''(2) = 36 > 0, so f' increases here.
3. f' has a local maximum at x=-1 and a local minimum at x=1, so f'' changes sign at both x=±1. This means f changes concavity at x=±1, so these are the inflection points of f.
4. Confirm directly: f''(x) = 12x² - 12 = 12(x² - 1), which crosses the x-axis at x=±1 and changes sign at both points, matching our analysis from f'.

Exam 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.

4. Sketching f From a Given Graph of f'

A very common FRQ problem gives you the graph of f' (often piecewise linear, for easy analysis) and an initial condition f(a) = k, and asks you to sketch f or identify its key features. The step-by-step process for this is: 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 max, min, or neither; 3) Split intervals further by local extrema of f' (these are inflection points of f), 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) + ∫ₐᵇ f'(x) dx, which simplifies to adding areas of geometric shapes when f' is piecewise linear.

Worked Example

Problem: 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. Write the equation of f': slope is (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: f(2) = f(0) + area under f' from 0 to 2 = 1 + (base * height)/2 = 1 + (2*4)/2 = 5, so the local maximum is at (2, 5).
3. Find concavity: f''(x) is the slope of f' = -2, which is always negative, so f is always concave down, with no inflection points.
4. Sketch: Start at (0, 1), increase with decreasing slope to (2, 5), then decrease with increasingly negative slope, remaining concave down everywhere, matching all features.

Exam 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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Looking for x-intercepts on f' to find inflection points of f. Why: 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 move: Memorize and write this on scratch paper for each problem: x-intercepts of f' = critical points of f; x-intercepts of f' / local extrema of f' = inflection points of f.
• Wrong move: Assuming that because f''(c) = 0, f must have an inflection point at x=c. Why: Students forget that inflection points require a sign change of f'', not just a zero value. For example, f(x) = x⁴ has f''(0) = 0 but no sign change, so no inflection point. Correct move: Always check that f'' changes sign around x=c after finding f''(c)=0 before confirming an inflection point.
• Wrong move: Swapping the sign of f' when working from f: marking f' negative when f is increasing. Why: Time pressure on the exam leads to flipped relationships when working backwards. Correct move: Write the base rule "f increasing → f' positive, f decreasing → f' negative" at the top of your scratch paper for matching problems before you start.
• Wrong move: Drawing f with an inflection point at the x-intercept of f' when sketching from f'. Why: This comes from confusing critical points and inflection points under time pressure. Correct move: 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 move: Claiming a cubic f must have a cubic f'. Why: Students forget that differentiation lowers the degree of a polynomial by 1. Correct move: For polynomial matching problems, first check the degree of f to eliminate any options for f' that have the wrong degree.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

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) $f^{\prime }(3)=0$ and $f^{\prime \prime }(-1)>0$ B) $f^{\prime }(3)=0$, $f^{\prime \prime }(-1)=0$, and $f^{\prime \prime }$ changes from negative to positive at $x=-1$ C) $f^{\prime }(3)>0$, $f^{\prime \prime }(-1)=0$, and $f^{\prime \prime }$ changes from negative to positive at $x=-1$ D) $f^{\prime }(3)=0$, $f^{\prime \prime }(-1)=0$, and $f^{\prime \prime }$ changes from positive to negative at $x=-1$

Worked Solution: Any local extremum of a differentiable function f occurs where $f^{\prime }(x)=0$, so $f^{\prime }(3)=0$, which eliminates option C immediately. A concavity change from down to up means f has an inflection point at $x=-1$, so $f^{\prime \prime }$ changes from negative (concave down) to positive (concave up) at $x=-1$, and $f^{\prime \prime }(-1)=0$ for twice-differentiable f. Option A is wrong because it does not include the zero value and sign change, and option D reverses the direction of the sign change. The correct answer is B.

Question 2 (Free Response)

The graph of $f^{\prime }(x)$, the first derivative of a twice-differentiable function f, is an upper semicircle centered at $(2,0)$ with radius 2, defined on $[0,4]$. It is 0 at $x=0$ and $x=4$, with a local maximum at $(2,2)$. You are given $f(0)=1$. (a) Identify the x-coordinate of any inflection points of f on $(0,4)$. Justify your answer. (b) Classify the critical point of f at $x=4$ as a local maximum, local minimum, or neither. Justify your answer. (c) Find the value of $f(4)$, given the area of a semicircle with radius 2 is $2\pi$.

Worked Solution: (a) Inflection points of f occur where $f^{\prime \prime }$ changes sign, which is where $f^{\prime }$ changes from increasing to decreasing or vice versa. $f^{\prime }$ increases from $x=0$ to $x=2$, then decreases from $x=2$ to $x=4$, so $f^{\prime }$ has a local maximum at $x=2$, meaning $f^{\prime \prime }$ changes from positive to negative at $x=2$. The only inflection point of f on $(0,4)$ is $\boxed{x=2}$. (b) $f^{\prime }(4)=0$, so $x=4$ is a critical point. For all $x<4$ near 4, $f^{\prime }(x)>0$ (the entire semicircle is above the x-axis except endpoints), so f is increasing for all $x<4$ near 4. There is no sign change of $f^{\prime }$ at $x=4$ on the interval $[0,4]$, so $x=4$ is $\boxed{\text{neither a local maximum nor minimum}}$. (c) By the Fundamental Theorem of Calculus, $f(4)=f(0)+{\int }_0^4{f}^{\prime }(x)dx=1+2\pi$, so $\boxed{f(4)=1+2\pi }$.

Question 3 (Application / Real-World Style)

The height of a projectile $t$ seconds after launch is $h(t)$, measured in meters. The graph of $h^{\prime }(t)$, the vertical velocity of the projectile, has a local maximum at $t=1$ second and a local minimum at $t=4$ seconds. $h^{\prime }(t)$ is positive for $0\le t<6$ and negative for $t>6$. When does the height graph have inflection points, and what do they mean in the context of the projectile's motion?

Worked Solution: Inflection points of $h(t)$ occur at the local extrema of $h^{\prime }(t)$, so inflection points occur at $\boxed{t=1}$ second and $\boxed{t=4}$ seconds. $h^{\prime \prime }(t)$ is the vertical acceleration of the projectile. At $t=1$, acceleration changes from positive to negative, meaning the projectile's velocity stops increasing and starts decreasing—this corresponds to the projectile exiting the cannon and gravity starting to slow its upward motion. At $t=4$, acceleration changes from negative to positive, meaning velocity stops decreasing and starts increasing (in the downward direction) after the projectile reaches its peak height.

7. Quick Reference Cheatsheet

8. 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 the next unit on integration, where you will repeatedly sketch antiderivative graphs from derivative graphs, using the same area calculation and feature mapping you practiced here.

• Optimization
• Curve Sketching
• Antiderivatives
• Fundamental Theorem of Calculus