Sketching Graphs of Functions and Their Derivatives: Study Guide (Calculus AB 5.8)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

What a sketch has to show

Topics 5.3 to 5.7 each found one feature of a graph. Sketching puts them together. Information about f can arrive in three forms: a formula (analytical), a graph of f′ or f″ (graphical) or a table of values (numerical). The connections are the same in every form.

Feature of the graph of f	What f′ does	What f″ does
f increasing	f′ > 0	—
f decreasing	f′ < 0	—
relative maximum at c	f′ changes from + to − at c	often f″(c) < 0
relative minimum at c	f′ changes from − to + at c	often f″(c) > 0
concave up	f′ increasing	f″ > 0
concave down	f′ decreasing	f″ < 0
point of inflection at c	f′ has a relative max or min at c	f″ changes sign at c
horizontal tangent at c	f′(c) = 0	—

Two warnings sit inside this table. A zero of f′ is only an extremum if f′ changes sign there. A zero of f″ is only a point of inflection if f″ changes sign there.

The sign-chart method

For a formula, follow these steps.

1. State the domain. Note any asymptotes or points where f is undefined.
2. Find f′. Mark where f′ = 0 or f′ does not exist (critical points).
3. Find f″. Mark where f″ = 0 or f″ does not exist.
4. Put all marked x-values on one number line. Test the sign of f′ and of f″ in each interval.
5. Read off the shape in each interval: rising or falling, cup or cap.
6. Calculate f at every marked x-value. Plot these points, then join them with the right shape.

There are four basic shapes, one for each sign combination:

f′	f″	Shape of f
+	+	rising, getting steeper
+	−	rising, flattening out
−	+	falling, flattening out
−	−	falling, getting steeper

Worked example 1: sketching from a formula

Question. Sketch the graph of f(x) = x⁴ − 4x³ + 16x. Label all relative extrema and points of inflection.

1. Domain. f is a polynomial: all real numbers, no asymptotes.
2. First derivative. f′(x) = 4x³ − 12x² + 16 = 4(x³ − 3x² + 4). Since x = −1 makes x³ − 3x² + 4 zero, (x + 1) is a factor: x³ − 3x² + 4 = (x + 1)(x − 2)². So f′(x) = 4(x + 1)(x − 2)². Critical points: x = −1 and x = 2.
3. Second derivative. f″(x) = 12x² − 24x = 12x(x − 2). Zeros: x = 0 and x = 2.
4. Sign chart. The marked values are −1, 0 and 2.

Interval	x < −1	−1 < x < 0	0 < x < 2	x > 2
sign of f′	−	+	+	+
sign of f″	+	+	−	+
shape of f	falling, flattening	rising, steepening	rising, flattening	rising, steepening

5. Read the features.
   • f′ changes from − to + at x = −1: relative minimum. It is also the absolute minimum, because f decreases before it and increases everywhere after it.
   • At x = 2, f′ = 0 but f′ is positive on both sides: no extremum.
   • f″ changes sign at x = 0 (+ to −) and at x = 2 (− to +): points of inflection at both.
6. Values. f(−1) = 1 + 4 − 16 = −11. f(0) = 0. f(2) = 16 − 32 + 32 = 16.

Check. The sketch matches every column of the sign chart: falling then rising, with the bend switching at x = 0 and again at x = 2.

Worked example 2: sketching f from the graph of f′

Often you are given the graph of f′, not f. Read it with two questions:

• Is f′ above or below the x-axis? That gives where f increases or decreases.
• Is f′ going up or down? That gives where f is concave up or concave down.

Question. The top panel of Figure 2 shows the graph of f′ for −2 ≤ x ≤ 4. It crosses the x-axis at x = −1 and x = 3 and has its highest point at (1, 4). You are told f(−1) = 0, f(1) = 16/3 and f(3) = 32/3. Describe f and sketch it.

1. Signs of f′. f′ < 0 for −2 ≤ x < −1, f′ > 0 for −1 < x < 3, f′ < 0 for 3 < x ≤ 4. So f decreases, then increases, then decreases.
2. Extrema. At x = −1, f′ crosses from below to above the axis (− to +): relative minimum. At x = 3, f′ crosses from above to below (+ to −): relative maximum.
3. Concavity. f′ is increasing for x < 1 and decreasing for x > 1. So f is concave up on −2 < x < 1 and concave down on 1 < x < 4.
4. Inflection. f′ has its maximum at x = 1, so the concavity of f changes there: point of inflection at x = 1. Also, f′(1) = 4 is the steepest slope of f on the interval.
5. Sketch. Plot (−1, 0), (1, 16/3) and (3, 32/3). Join them: a cup through the minimum, rising and steepening up to x = 1, then rising and flattening into a cap at x = 3, then falling.

A common trap. The highest point of the f′ graph is not a maximum of f. It is where f is rising fastest, which is a point of inflection.

Worked example 3: conclusions from a table

Question. f is twice differentiable. The table gives values of f′. You are also told that f′ is decreasing on 0 < x < 3 and increasing on 3 < x < 5.

x	0	1	2	3	4	5
f′(x)	5	2	0	−1	0	3

Describe the behaviour of f on 0 < x < 5.

1. Sign of f′. Because f′ is decreasing on (0, 3) and f′(2) = 0, f′ is positive just before x = 2 and negative just after. Because f′ is increasing on (3, 5) and f′(4) = 0, f′ is negative just before x = 4 and positive just after.
2. Extrema. Relative maximum at x = 2 (f′ changes + to −). Relative minimum at x = 4 (f′ changes − to +).
3. Concavity. f′ decreasing on (0, 3): f concave down. f′ increasing on (3, 5): f concave up. Point of inflection at x = 3.
4. Steepness. f′(0) = 5 and f′(5) = 3, so f is rising more steeply at x = 0 than at x = 5.

Why the extra sentence matters. A table alone shows only a few values. Without knowing how f′ behaves between them, you cannot be sure that f′ has no other zeros. Always use the information given about the whole interval, and say so in your justification.

Going the other way: sketching f′ from f

The same table works in reverse. Given the graph of f:

• Where f has a horizontal tangent (a peak, a trough or a flat inflection), f′ crosses or touches the x-axis.
• Where f rises, the graph of f′ is above the axis; where f falls, it is below.
• Where f is concave up, f′ is increasing; where f is concave down, f′ is decreasing.
• At a point of inflection of f, f′ has a peak or a trough.
• At a corner or cusp of f, f′ is undefined, and its graph has a gap or jump there.

For Worked example 1, the graph of f′ is below the axis for x < −1, crosses at −1, rises to a peak at x = 0, falls to touch the axis at x = 2 without crossing, and then rises again.

Common misconceptions

• Reading the graph of f′ as if it were f. A peak on the f′ graph is a point of inflection of f, not a maximum.
• “f′ = 0, so there is a maximum or minimum.” Not if f′ touches the axis without changing sign, as at x = 2 in Worked example 1.
• “f″ = 0, so there is a point of inflection.” Only if f″ changes sign.
• Taking concavity from the sign of f′. Concavity comes from whether f′ is increasing or decreasing, not whether it is positive.
• Joining key points with straight lines. Each interval needs its correct bend: a cup or a cap.
• Leaving key points unlabelled. A sketch should show the coordinates of every extremum and inflection point.
• Concluding from a table alone. You need information about f′ between the listed values.

Where this leads

Sketching ties Unit 5 together. In Topic 5.9, Connecting a Function, Its First Derivative, and Its Second Derivative, you match the graphs of f, f′ and f″ and read features of one from another. The same reasoning returns in Unit 6, where you rebuild f from the graph of f′ using areas. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.