A. Rate of Change · Derivatives and Differentiation Rules

Connecting the Graphs of a Function and Its Derivative

A2.1 A2.2About 65 minutes

Learning goals

I can describe how the graph of f′ shows where f increases, decreases and turns.
I can match the graph of a function with the graph of its derivative and explain my choice.
I can sketch a possible graph of f from the graph of f′, and explain why there are many.
I can connect the symmetry of a polynomial with the symmetry of its derivative.

Words to know

stationary point
A point where f′(x) = 0, so the tangent is horizontal. It may be a local max, a local min, or neither.

steepness
How fast the graph rises or falls; measured by the size of f′(x).

even function
A function with f(−x) = f(x); its graph is symmetric in the y-axis.

odd function
A function with f(−x) = −f(x); its graph has rotational symmetry about the origin.

Two graphs, one story

A function and its derivative are like a hiking trail and its steepness chart. The trail map shows the height; the steepness chart shows how hard each part of the climb is. If you know one, you can say a lot about the other.

In this lesson you will read the two graphs side by side, match functions with their derivatives, and sketch one from the other. (In the unit Curve Sketching: Connecting f, f′ and f″ you will go further, using the second derivative to describe how the graph bends.)

The connections

Each line pairs a feature of the graph of f with the matching feature of the graph of f′:

f is increasing ⟷ f′ is positive (above the x-axis).
f is decreasing ⟷ f′ is negative (below the x-axis).
f has a horizontal tangent (often a local max or min) ⟷ f′ has a zero (x-intercept).
f is steep ⟷ f′ is far from the x-axis. f is nearly flat ⟷ f′ is close to the x-axis.
f is a straight line ⟷ f′ is constant (horizontal).
f has a corner ⟷ f′ is undefined there (it jumps).

The y-values of f do not show up on f′ at all. Moving f up or down changes nothing on f′.

A cubic f (top) and its derivative f′ (bottom, dashed), lined up on the same x-scale. The shaded intervals are where f′ is above the axis and f is rising. Each turning point of f sits directly above a zero of f′, and the lowest point of f′ (x = 1) is where f falls most steeply.

Vertical guide lines

Draw light vertical lines through every turning point of f before you sketch f′. The zeros of f′ must sit on those lines. Then fill in the sign of f′ between them (+ where f rises, − where it falls).

Worked example

Example 1: Reading the pair of graphs

Use the graphs of f(x) = x³/3 − x² − 3x + 2 and f′(x) = x² − 2x − 3 in the figure. (a) Where does f have its turning points? (b) Where is f decreasing? (c) What is the slope of f at x = 1, and why is it special?

(a) f′(x) = x² − 2x − 3 = (x − 3)(x + 1), which is 0 at x = −1 and x = 3. The sign of f′ changes from + to − at −1 (local max) and from − to + at 3 (local min).
f(−1) = −1/3 − 1 + 3 + 2 = 11/3 and f(3) = 9 − 9 − 9 + 2 = −7.
(b) f′ < 0 between its zeros, so f decreases for −1 < x < 3.
(c) f′(1) = 1 − 2 − 3 = −4. This is the vertex (lowest point) of the parabola y = f′(x), so x = 1 is where f is falling most steeply.

(a) Local max (−1, 11/3), local min (3, −7). (b) −1 < x < 3. (c) Slope −4, the steepest downhill slope of f.

Matching puzzle for Example 2. Each graph in the top row is a function; each graph in the bottom row is the derivative of one of them. All six use the same scales, with ticks at x = −1 and x = 1.

Worked example

Example 2: Matching functions with derivatives

Match each function A, B and C in the matching-puzzle figure with its derivative 1, 2 or 3.

A is a parabola opening up with its vertex at x = 0. It decreases, then increases, so its derivative is negative, then zero at 0, then positive: a line through the origin with positive slope. That is graph 3.
B decreases, rises between x = −1 and x = 1, then decreases. Its derivative is negative, then positive between −1 and 1, then negative: a parabola opening down with zeros at ±1. That is graph 1.
C rises everywhere and is flat only at x = 0. Its derivative is positive everywhere except a zero at 0: a parabola opening up that touches the axis at the origin. That is graph 2.

A → 3, B → 1, C → 2. (The equations are A: x² − 2, B: −x³/3 + x, C: x³/3, with derivatives 2x, −x² + 1 and x².)

Worked example

Example 3: From f′ back to f

The derivative of a function g is the line g′(x) = 2x − 4. Sketch a possible graph of g. How many correct answers are there?

g′(x) < 0 for x < 2, so g decreases there; g′(x) > 0 for x > 2, so g increases.
g′ changes from − to + at x = 2: g has a local minimum at x = 2.
g′ is linear, so g is quadratic: a parabola opening up with its vertex at x = 2, for example g(x) = x² − 4x.
Shifting this parabola up or down does not change any slope, so g(x) = x² − 4x + C works for any constant C.

Any upward parabola with its vertex on the line x = 2, such as g(x) = x² − 4x + C. There are infinitely many correct graphs, all vertical translations of each other.

Worked example

Example 4: Symmetry

f(x) = x⁴ − 2x² is an even function. Its derivative is f′(x) = 4x³ − 4x. Show that f′ is odd, and explain why with tangent slopes.

f′(−x) = 4(−x)³ − 4(−x) = −4x³ + 4x = −(4x³ − 4x) = −f′(x), so f′ is odd.
Why: the graph of an even function is a mirror image in the y-axis. The mirror image of a tangent that rises to the right is a tangent that falls to the right, with the opposite slope.
So the slope at −x is the negative of the slope at x.

f′(−x) = −f′(x): the derivative of an even polynomial is odd. (Similarly, the derivative of an odd polynomial is even, as with x³ and 3x².)

Just for fun

A function walks into a mirror shop and asks for a reflection that shows how it really feels. The shopkeeper hands it a derivative: "This one shows your ups and downs."

Hands-on activity

Trail map and steepness chart

You need: a long piece of string or a phone charging cable, a table, two sheets of paper, a pencil and tape.

Lay the string on a sheet of paper in a gentle wave with two hills and one valley. Tape it down and trace it. This is your graph of f.
Directly under it on the second sheet, draw an x-axis the same length. Draw dashed vertical lines down from each hilltop and the valley bottom.
Mark a dot on the axis where each dashed line meets it: these are zeros of f′.
Between the zeros, decide whether f′ is above or below the axis (rising or falling trail).
Find the steepest part of each climb and descent on f. Make f′ reach its highest or lowest value there. Join everything smoothly.
Swap with a family member: they draw an f′, and you sketch an f to match.

What to look for: f′ should have three zeros (two from the hilltops, one from the valley), and the signs should read +, −, +, − from left to right. Any vertical position for the reconstructed f is correct.

Your turn

f has a local maximum at x = 4. What does the graph of f′ do at x = 4?

f′ is negative for all x. Describe the graph of f.

f is a straight line with slope −3. Describe the graph of f′.

The graph of f′ crosses the x-axis at x = −2 (from − to +) and at x = 5 (from + to −). Describe f.

f(x) = x² + 1 and g(x) = x² − 6. How do their derivative graphs compare?

At x = 2 the graph of f is very steep and rising. What can you say about f′(2)?

Is the derivative of the odd function f(x) = x³ − 3x even, odd or neither?

More practice: warm-up

Where f is increasing, f′ is …

A turning point of f corresponds to what on f′?

If f is a parabola, what shape is f′?

If f is constant, what is f′?

Does f′ show how high f is?

More practice: on your own

f is a cubic with a local max at x = −3 and a local min at x = 1. Sketch f′ in words, including its zeros, its shape and where it is lowest.

The graph of f′ is a parabola opening down with zeros at x = 0 and x = 6. Describe the graph of f.

The graph of f′ is the horizontal line y = 0.5. Find a possible equation of f and say what the graph looks like.

For f(x) = |x + 1|, describe f′.

Graph P is a parabola with vertex (2, 5) opening down. Graph Q is the line y = −2x + 4. Could Q be the derivative of P? Explain.

Challenge

A quartic f has local minimums at x = −2 and x = 2 and a local maximum at x = 0. What is the degree of f′, what are its zeros, and what are its signs?

Two different functions have the derivative graph y = 3x². One passes through (0, 0) and the other through (0, 4). Give their equations.

A smooth function f has f′(x) > 0 for x < 1, f′(1) = 0 and f′(x) > 0 for x > 1. f(1) = 3. Sketch f in words and name a possible equation.

Check your understanding

Quiz

Where the graph of f′ is above the x-axis, the graph of f is:

f has turning points at x = −4 and x = 2. Where must f′ be zero?

The derivative of a parabola opening up with vertex (−3, 1) is:

Which change to f leaves f′ unchanged?

f′ is a parabola that touches the x-axis only at x = 1 and is positive elsewhere. At x = 1, f has:

The derivative of an even polynomial is:

For parents and teachers

Big idea (A2.1, A2.2): the graph of f′ records the steepness of the graph of f. Rising parts of f give positive f′, falling parts give negative f′, and turning points give zeros of f′.

How to help: Draw a quick graph and ask your teen to sketch its derivative underneath, or the reverse. Ask "Where are the zeros? Why?" and "Where is f steepest?" Praise answers that use the words slope, tangent and sign.

Common difficulties:

Copying the shape of f instead of its slopes (for example, drawing a hill for f′ under a hill of f).
Putting the zeros of f′ under the x-intercepts of f instead of under the turning points.
Thinking there is only one f for a given f′; any vertical translation also works.