Wemazi

Topic 5.8

Match a function with its derivative

Builds on reading concavity from a derivative graph. 7 minutes.

Which graph is the slope?

The top graph shows \(f\), and one of the three candidates below it is \(f'\). You find it by comparing features, such as where each graph is 0 and where it is above the axis.

Press Show the guides. The dashed lines mark \(x = -1\) and \(x = 3\), where \(f\) has a horizontal tangent, so \(f'\) must be 0 at both inputs. C is 3 at \(x = -1\), so C cannot be \(f'\), and a red ring marks the failure.

Press Move to \(x = 0\), where the tangent of \(f\) slopes down. Its slope is negative, but B is 3, so B cannot be \(f'\).

A is \(-3\) at \(x = 0\), the slope of the tangent there, so A is \(f'\).

Check one feature at a time

Press Check zeros. The derivative is 0 where \(f\) has a horizontal tangent, so a candidate that is not 0 there fades.

Press Check signs. The curve of \(f\) is green where it rises and red where it falls. Each candidate is green above its axis and red below it. The derivative is positive where \(f\) rises and negative where \(f\) falls, so a candidate with the wrong sign fades.

Press Check bends. One dot marks where \(f\) changes from concave down to concave up. The other marks the lowest point of A there, because \(f'\) falls while \(f\) is concave down and rises while \(f\) is concave up.

One failed feature rejects a candidate. The graph that passes every test is \(f'\).

Notation

  1. At each input \(c\) where \(f\) has a horizontal tangent, the graph of \(f'\) must be 0.

    \[f'(c) = 0\]

  2. Compare signs. Where \(f\) rises, the graph of \(f'\) is above the axis, and where \(f\) falls, it is below.

    \[\begin{aligned} \cg{f' > 0}&\ \text{where } f \text{ rises} \\ \cr{f' < 0}&\ \text{where } f \text{ falls} \end{aligned}\]

  3. Compare size. A steeper tangent of \(f\) gives a value of \(f'\) farther from the axis. A is lowest where \(f\) falls most steeply.

  4. Compare bends. Where the graph of \(f'\) rises, \(f\) is concave up. Where it falls, \(f\) is concave down.

  5. Reject a candidate as soon as one feature fails. Accept only a graph that passes every test on the whole interval.

When it fails

Move the graph of \(f\) up or down. Its tangent keeps its slope, so A still matches.

Two functions that differ by a constant have the same derivative, so a graph of \(f'\) fixes the shape of \(f\), not its height.

Example 1: a cosine curve

The graph of \(f(x) = \cos x\) is given on \([0, 2\pi]\). Which is \(f'\): A = \(\sin x\), B = \(-\sin x\), or C = \(\cos x\)?

  1. The graph of \(f\) has horizontal tangents at \(x = 0\), \(\pi\) and \(2\pi\), so \(f'\) must be 0 at these inputs.

  2. C is 1 at \(x = 0\), where \(f\) is flat, so C cannot be \(f'\).

    \[\cos 0 = 1 \neq 0\]

  3. A and B are 0 at the three inputs, so compare signs. The graph of \(f\) falls between 0 and \(\pi\), so \(f'\) is negative there.

  4. At \(x = \tfrac{\pi}{2}\), A is 1 and B is \(-1\). A is positive where \(f\) falls, so A cannot be \(f'\), and B is \(f'\).

    \[f'(x) = -\sin x\]

Example 2: a cubic

The graph of \(f(x) = \dfrac{x^3}{3} - x\) is given. Which is \(f'\): A = \(x^2 - 1\), B = \(x^3 - x\), or C = \(1 - x^2\)?

  1. The graph of \(f\) has horizontal tangents at \(x = -1\) and \(x = 1\), so \(f'\) is 0 at these inputs.

  2. A tempting wrong move: The graph of \(f\) is S-shaped, so \(f'\) must be S-shaped too. The candidate is B.

    \[B(x) = x^3 - x\]

    B is 0 at three inputs, \(-1\), 0 and 1. At \(x = 0\) the tangent of \(f\) slopes down, so \(f'\) is not 0 there.

  3. Keep A and C, the candidates that are 0 only at \(-1\) and 1.

    \[\begin{aligned} x^2 - 1 = 0 \ &\text{and}\ 1 - x^2 = 0 \\ &\Rightarrow\ x = \pm 1 \end{aligned}\]

  4. At \(x = 0\), the tangent of \(f\) has slope \(-1\). A is \(-1\) there, and C is 1, so A is \(f'\).

    \[A(0) = -1 \qquad C(0) = 1\]

Practice

The graph of \(f(x) = \sin 2x\) is given on \([0, \pi]\). Which is \(f'\): A = \(\cos 2x\), B = \(2\cos 2x\), or C = \(-2\cos 2x\)?

The candidates are 0 at the same inputs, so compare the slope of \(f\) at \(x = 0\) with each height there.

Solution

  1. The tangents of \(f\) are horizontal at \(\tfrac{\pi}{4}\) and \(\tfrac{3\pi}{4}\). All three candidates are 0 there, so the zeros do not help.

    \[\cos\tfrac{\pi}{2} = 0 \qquad \cos\tfrac{3\pi}{2} = 0\]

  2. The graph of \(f\) rises at \(x = 0\), so \(f'(0)\) is positive. C is \(-2\) at 0, so C cannot be \(f'\).

    \[C(0) = -2\]

  3. Near 0, the curve \(\sin 2x\) is close to \(2x\), so its tangent has slope 2. A is 1 at 0, and B is 2, so B is \(f'\).

    \[A(0) = 1 \qquad B(0) = 2\]

Questions

The top graph is \(f(x) = \tfrac{x^4}{4} - 2x^2\). Below it are three candidates for \(f'\).

  1. Which candidate do the horizontal tangents of \(f\) rule out?

    • C
    • A
    • B
    • None of them
    Answer

    C

  2. The point is at \(x = 3\), where \(f\) rises. Which candidate is \(f'\)?

    • A
    • B
    • C
    Answer

    A

  3. Let \(f(x) = x^2 - 6x + 1\) and \(g(x) = x^2 - 6x + 8\). Which statement about the graphs of \(f'\) and \(g'\) is correct?

    • The graphs of \(f'\) and \(g'\) are the same
    • The graph of \(g'\) is the graph of \(f'\) shifted up by 7
    • The graphs of \(f'\) and \(g'\) cross only at \(x = 3\)
    Answer

    The graphs of \(f'\) and \(g'\) are the same

Summary

You test each candidate against the horizontal tangents, direction, steepness and bends of \(f\), and one failed feature rejects it.

This works because the graph of \(f'\) records the slope of \(f\) at every input.

You use it when a problem gives a graph of \(f\) and asks which graph is \(f'\).

This is one of five skills in topic 5.8 of Which & Why: AP Calculus.