AP CALCULUS BC • ANALYTICAL APPLICATIONS OF DIFFERENTIATION

Sketching Graphs of Functions and their Derivatives

Translate between a function, its first derivative, and its second derivative using analytical and graphical reasoning.

Historical Context & Motivation

The ability to sketch the graph of a function from analytical information about its derivatives is one of the oldest and most powerful applications of calculus. Before graphing calculators or computer algebra systems existed, mathematicians and scientists relied on curve sketching as the primary tool for understanding how functions behave over their domains. The interplay between a function and its derivatives is not merely computational; it reveals the underlying geometry of change — where a quantity increases, where it reaches an extreme, and how sharply it bends. This interplay sits at the heart of the Analytical Applications of Differentiation unit on the AP Calculus BC exam, and mastering it gives you the ability to reason about functions even when an explicit formula is not available.

1684
Leibniz Publishes His Calculus
Gottfried Wilhelm Leibniz introduced the notation dy/dx and formulated rules for differentiation, providing the symbolic framework for analyzing rates of change.
1696
L'Hôpital's Analyse des Infiniment Petits
The first calculus textbook systematically applied derivatives to locate maxima, minima, and inflection points of curves, formalizing curve-sketching techniques.
1748
Euler's Introductio in Analysin Infinitorum
Leonhard Euler unified algebraic and graphical perspectives, classifying functions by their analytic properties and the shapes of their graphs.
1797
Lagrange's Théorie des Fonctions Analytiques
Joseph-Louis Lagrange formalized the concept of the derivative as a derived function, establishing the second derivative test and concavity analysis still taught today.
1960s–present
Graphing Technology & Conceptual Understanding
With the advent of graphing calculators and CAS software, emphasis shifted from routine plotting to conceptual fluency — interpreting derivative graphs and translating between representations.

The central question this lesson addresses is deceptively simple: Given one of the three representations — f, f′, or f″ — how do you reconstruct the other two? Mastering the answer to this question prepares you for multiple-choice and free-response items that ask you to identify extrema, intervals of increase or decrease, concavity, and inflection points from derivative graphs — all without a calculator.

Core Principles & Definitions

Before sketching any graph, you need a rock-solid mental map of how a function and its derivatives relate to one another. Every feature of the graph of f — its hills, valleys, slopes, and curvature — can be read directly from the signs and values of f′ and f″. Conversely, every zero-crossing or sign change in the graph of f′ corresponds to a specific geometric event on the graph of f. The five foundational ideas below form the backbone of all graph-sketching analysis.

1

f′ > 0 ⇒ f is Increasing

When the first derivative is positive on an interval, the original function is strictly increasing on that interval. The graph of f rises from left to right.
2

f′ < 0 ⇒ f is Decreasing

When the first derivative is negative, f is strictly decreasing. The graph of f falls from left to right. A sign change in f′ from positive to negative signals a local maximum.
3

f′ = 0 or DNE ⇒ Critical Point

A critical point occurs where f′ equals zero or does not exist (and f is defined). Not every critical point is an extremum — the First or Second Derivative Test resolves this.
4

f″ > 0 ⇒ Concave Up

When the second derivative is positive, the graph of f bends upward (like a cup). This means f′ is increasing — the slope of f is getting steeper in the positive direction.
5

f″ Changes Sign ⇒ Inflection Point

An inflection point occurs where f changes concavity — from concave up to concave down or vice versa. This corresponds to a local extremum of f′.
KEY TAKEAWAY
Think of a function, its first derivative, and its second derivative as three levels of a control tower. The function f tells you where you are (position), f′ tells you how fast you're going (velocity), and f″ tells you whether you're speeding up or slowing down (acceleration). An engineer monitoring a rocket's altitude reads the same trio of instruments: altimeter, speedometer, and accelerometer. Zeros in the velocity graph correspond to altitude peaks or troughs; zeros in the acceleration graph correspond to the moments the rocket changes from accelerating to decelerating.

Visual Explanation — From f to f′ to f″

The diagram below shows three vertically stacked graphs — the original function f on top, its first derivative f′ in the middle, and its second derivative f″ at the bottom — all sharing a common x-axis. Study how the key features of f align vertically with the sign changes and zeros of f′ and f″. Vertical dashed lines connect the corresponding features across all three graphs.

The three graphs are stacked so that each x-value lines up vertically. Notice that where f has a local maximum (x ≈ 270), f′ crosses zero from positive to negative. Where f has an inflection point (x ≈ 370), f″ crosses zero. These vertical correspondences are the key to translating between representations.

The diagram makes the following correspondences explicit. Wherever f has a local extremum, the graph of f′ crosses the x-axis — it equals zero and changes sign. The direction of the sign change tells you whether the extremum is a max (positive to negative) or a min (negative to positive). Wherever f has an inflection point, f″ crosses the x-axis — concavity switches. Equivalently, f′ reaches a local extremum at that same x-value. Understanding this vertical alignment is perhaps the single most important skill tested on the AP exam's derivative-graph questions.

Mathematical Framework

The formal tools for graph sketching rest on three major results from calculus: the First Derivative Test, the Second Derivative Test, and the concavity characterization theorem. Together these theorems allow you to classify every critical point and determine the exact shape of the curve between critical points.

FIRST DERIVATIVE TEST
If f′ changes from + to − at x = c, then f(c) is a local maximum. If f′ changes from − to + at x = c, then f(c) is a local minimum.
Here c is a critical number where f′(c) = 0 or f′(c) does not exist, and f is continuous at c. The sign of f′ on either side of c determines the classification.
SECOND DERIVATIVE TEST
If f′(c) = 0 and f″(c) > 0, then f(c) is a local minimum. If f′(c) = 0 and f″(c) < 0, then f(c) is a local maximum. If f″(c) = 0, the test is inconclusive.
This test is often faster than the First Derivative Test because it requires evaluating f″ at a single point rather than analyzing sign changes on intervals. However, when f″(c) = 0 you must fall back to the First Derivative Test.
CONCAVITY THEOREM
f″(x) > 0 on (a, b) ⇒ f is concave up on (a, b) f″(x) < 0 on (a, b) ⇒ f is concave down on (a, b)
Concave up means the tangent line lies below the curve; concave down means the tangent line lies above the curve. Visually, concave up curves 'hold water' and concave down curves 'spill water.'
INFLECTION POINT CONDITION
x = c is an inflection point of f if f″ changes sign at x = c.
It is not sufficient for f″(c) to equal zero; the sign of f″ must actually change. For example, f(x) = x⁴ has f″(0) = 0 but x = 0 is not an inflection point because f″ does not change sign there.

When the AP exam presents you with the graph of f′ and asks about f, apply these results in reverse. The zeros of f′ are candidates for extrema of f. The sign of f′ tells you whether f is increasing or decreasing. The slope of f′ (which is f″) tells you the concavity of f. Extrema of f′ correspond to inflection points of f. These reverse readings are essential because the exam frequently gives you f′ rather than f.

Detailed Breakdown — The Sign Chart Method

A systematic approach to sketching or interpreting graphs involves constructing sign charts for f′ and f″. A sign chart partitions the domain at critical numbers and possible inflection points, then records the algebraic sign on each resulting interval. From two sign charts you can deduce the complete qualitative shape of f. The diagram below illustrates the process for f(x) = x³ − 3x² − 9x + 5, walking through the sign analysis and showing how each interval's behavior maps to the graph.

The sign chart for f′ identifies the critical numbers x = −1 and x = 3. The sign chart for f″ identifies the inflection point at x = 1. Combined, these charts give a complete qualitative picture of the graph: increasing and concave down on (−∞, −1), decreasing and concave down on (−1, 1), decreasing and concave up on (1, 3), and increasing and concave up on (3, ∞).
💡 AP Exam Tip
On the AP Calculus BC exam, you may be given the graph of f′ (not f) and asked to determine the number of relative extrema, intervals of concavity, or inflection points of f. Remember: x-intercepts of f′ are critical points of f, and local extrema of f′ are inflection points of f. Always check that f′ actually changes sign for a relative extremum, not just touches the axis.

Worked Example

Let us work through a complete graph-sketching analysis for the function f(x) = x⁴ − 4x³. We will find critical points, classify them, determine concavity, locate inflection points, and assemble the sketch.

Sketch the graph of f(x) = x⁴ − 4x³
1
Step 1 — Compute f′(x)Differentiate using the power rule: f′(x) = 4x³ − 12x². Factor: f′(x) = 4x²(x − 3).
f′(x) = 4x²(x − 3)
2
Step 2 — Find critical numbersSet f′(x) = 0: 4x²(x − 3) = 0 gives x = 0 (multiplicity 2) and x = 3. Both are in the domain of f, so both are critical numbers.
Critical numbers: x = 0, x = 3
3
Step 3 — First Derivative Test (sign chart for f′)Test values: f′(−1) = 4(1)(−4) = −16 < 0, so f is decreasing on (−∞, 0). f′(1) = 4(1)(−2) = −8 < 0, so f is still decreasing on (0, 3). f′(4) = 4(16)(1) = 64 > 0, so f is increasing on (3, ∞). At x = 0 there is no sign change (negative on both sides), so x = 0 is not a relative extremum. At x = 3, f′ changes from negative to positive, so x = 3 is a local minimum.
Local min at x = 3, f(3) = 81 − 108 = −27
4
Step 4 — Compute f″(x) and find inflection pointsf″(x) = 12x² − 24x = 12x(x − 2). Set f″(x) = 0: x = 0 and x = 2. Check sign changes: f″(−1) = 12(−1)(−3) = 36 > 0, f″(1) = 12(1)(−1) = −12 < 0, f″(3) = 12(3)(1) = 36 > 0. The sign of f″ changes at both x = 0 and x = 2, so both are inflection points.
Inflection points: (0, 0) and (2, −16)
5
Step 5 — Assemble the sketchKey coordinates: f(0) = 0, f(2) = 16 − 32 = −16, f(3) = −27. The curve enters from the upper-left (since the leading coefficient is positive and the degree is even), decreases through (0, 0) — concavity changes from up to down — continues decreasing through (2, −16) where concavity changes back to up, reaches its minimum at (3, −27), then increases toward +∞. The y-intercept is (0, 0).
Decreasing on (−∞, 3), increasing on (3, ∞); concave up on (−∞, 0) ∪ (2, ∞), concave down on (0, 2).
⚠️ Common Pitfall
Notice that x = 0 is a critical number (f′(0) = 0) but is not a local extremum because f′ does not change sign there. Many students assume every critical number yields a max or min — that is false. Always verify a sign change in f′. This is also why the Second Derivative Test is inconclusive when f″(c) = 0, as it is here at x = 0.

First Derivative Test vs. Second Derivative Test

Both the First Derivative Test (FDT) and the Second Derivative Test (SDT) serve to classify critical points, but they differ in approach, computational cost, and reliability. The table below highlights these distinctions so you can decide which test to apply in a given situation on the AP exam.

Comparison of extremum classification tests
FeatureFirst Derivative TestSecond Derivative Test
What you evaluateSign of f′ on intervals surrounding the critical numberValue of f″ at the critical number
ConclusivenessAlways conclusive — directly checks sign changeInconclusive when f″(c) = 0
Computational effortRequires testing f′ at sample points in each intervalRequires computing f″ and evaluating at one point
Works when f′(c) DNEYes — applicable at cusps, cornersNo — requires f′(c) = 0
Best used whenf′ is easy to factor or when given the graph of f′f″ is quick to compute and evaluate
KEY TAKEAWAY
Use the SDT as your quick first pass — it is like a fast diagnostic scan at a hospital. If the scan returns a clear result (f″(c) ≠ 0), you have your answer immediately. If the scan is inconclusive (f″(c) = 0), you run the more thorough test — the FDT — which examines the sign of f′ on a full interval, analogous to an in-depth MRI. On the AP exam, always start with the SDT when the algebra is clean, but have the FDT ready as your fallback.

Connections to Integration & Advanced Topics

Graph sketching from derivative information connects directly to several advanced topics on the AP Calculus BC exam. The Fundamental Theorem of Calculus provides the reverse direction: if you are given f′ (or f″) as a function, you can recover f (or f′) through integration. Accumulation functions, defined as F(x) = ∫ₐˣ f(t) dt, are analyzed using the same increasing/decreasing and concavity reasoning — F′(x) = f(x) by the FTC, so the graph of f acts as the 'derivative graph' for the accumulation function F.

How graph-sketching reasoning extends to BC-specific topics
This Lesson's ConceptAdvanced BC Extension
Sign chart for f′ → increasing/decreasing fIf f = F′ for an accumulation function F, the graph of f tells you where F increases and decreases (FTC Part 1)
Inflection points of f from f″ sign changesInflection points of parametric curves: analyze d²y/dx² = [d/dt(dy/dx)] / (dx/dt)
Local extrema from critical pointsOptimization problems in polar/parametric contexts; extrema of r(θ) or arc-length functions
Concavity and tangent-line approximationsError analysis for Euler's method: concavity determines whether Euler's approximation under- or overestimates

The core skill of reading a derivative graph and deducing properties of the original function is arguably the single most transferable skill in calculus. Whether you encounter differential equations, Taylor polynomial error bounds, or convergence tests, the ability to reason about the sign and behavior of a derivative will serve you throughout the course.

Practice Problems

1
The graph of f′ is continuous and has x-intercepts at x = −2 and x = 4. On the interval (−∞, −2), f′(x) > 0. On the interval (−2, 4), f′(x) < 0. On the interval (4, ∞), f′(x) > 0. Which of the following must be true about f?
2
Let f(x) = 2x³ − 9x² + 12x − 4. Find all intervals on which f is increasing.
3
The function g is twice differentiable, and the graph of g′ is a downward-opening parabola with vertex at (3, 5) and x-intercepts at x = 1 and x = 5. Determine the x-coordinate of the inflection point of g and identify the intervals on which g is concave up and concave down.
PROBLEM 4APPLIED
A particle moves along the x-axis with velocity v(t) = t³ − 6t² + 9t for t ≥ 0. (a) Find all times t at which the particle changes direction. (b) On what intervals is the particle's speed increasing? (c) Find the position x(t) if x(0) = 2, and determine the total distance traveled from t = 0 to t = 4.
PROBLEM 5CRITICAL THINKING
Let f be a twice-differentiable function on (−∞, ∞) with exactly two inflection points. Suppose f′(x) > 0 for all x, f″(−1) = 0, f″(3) = 0, and f″(x) > 0 on (−∞, −1) ∪ (3, ∞). Sketch a possible graph of f′ consistent with this information, clearly indicating any local extrema of f′ and the concavity behavior they imply for f. Justify why f cannot have any relative extrema.

Summary

Sketching graphs from derivative information is built on a small number of powerful correspondences. When f′ > 0, f is increasing; when f′ < 0, f is decreasing. Critical points occur where f′ = 0 or f′ does not exist, and the First Derivative Test classifies them by checking sign changes in f′. The Second Derivative Test provides a quicker classification when f″(c) ≠ 0 at a critical number c. Concavity is determined by the sign of f″: positive means concave up, negative means concave down. Inflection points occur where f″ changes sign, which corresponds to local extrema of f′.

The most critical AP exam skill is translating between graphs: given the graph of f′, read off the zeros of f′ for extrema of f, the sign of f′ for increasing/decreasing behavior, and the increasing/decreasing behavior of f′ (i.e., the slope of the f′ graph) for concavity of f. This three-level perspective — position, velocity, acceleration — unifies all of the analytical applications of differentiation and extends naturally to accumulation functions, parametric analysis, and differential equations encountered throughout the BC curriculum.

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