Wemazi

Topic 5.4

Test the derivative on both sides

Builds on listing the candidates for local extrema. 6 minutes.

The signs on both sides

The upper graph shows \(f(x) = 3x^5 - 5x^3\), green where it rises and red where it falls. Under the graph of \(f'\), a line will record its sign, green for positive and red for negative.

The tangent is horizontal at \(x = -1\), \(x = 0\) and \(x = 1\), so these are the candidates.

Press Go to \(x = -1\). The curve is green before \(-1\) and red after it, so \(f'\) is positive, then negative.

Do the same at 0 and at 1. At 0 the curve is red on both sides, so \(f'\) is negative before and after. At 1 it changes from red to green.

You read the signs of \(f'\) on the two sides of a candidate. The value \(f'(c)\) does not help, because it is 0 or missing at every candidate.

One test point gives the sign of its side

The orange point marks a candidate \(c\). At the same distance \(d\) on each side, orange dots mark \(f\) and \(f'\). The green plus or red minus beside the lower dot is the sign of \(f'\) there.

A cut point is an input where \(f'\) is 0 or does not exist, a dot on the line under the graphs. Between two cut points \(f'\) exists and is never 0, so its sign cannot change there.

Change \(d\). The signs do not change, because both dots stay between \(c\) and the next cut point on their side.

This means that one dot gives the sign of \(f'\) on its whole side.

Notation

  1. Start with a candidate \(c\), an input where \(f'(c)\) is 0 or does not exist.

  2. Choose a test point \(a\) between \(c\) and the next cut point on the left. Find the sign of \(f'(a)\).

    \[a < c:\quad \cy{f'(a)}\]

  3. Choose a test point \(b\) between \(c\) and the next cut point on the right. Find the sign of \(f'(b)\).

    \[b > c:\quad \cy{f'(b)}\]

  4. Write the two signs from left to right, here at each of the three candidates. Four patterns are possible: two change sign and two do not.

    \[\begin{gathered} (\cg{+},\cr{-}),\quad (\cr{-},\cg{+}) \\ (\cg{+},\cg{+}),\quad (\cr{-},\cr{-}) \end{gathered}\]

When it fails

Press Test the flat spot, the third candidate, where the tangent is horizontal but the curve does not turn. There \(f'\) touches 0 and stays positive, so both dots show a plus.

Press the second button to move the left dot past the next cut point. There \(f'\) is negative, because it changes sign at that cut point. The dots now read minus, then plus, as if the curve turned. A test point must lie between \(c\) and the next cut point.

Example 1: two candidates

Find the signs of \(f'\) on both sides of each candidate of \(f(x) = \dfrac{x^3}{3} - 2x^2 + 3x\).

  1. Factor \(f'\) to find the candidates, 1 and 3.

    \[\begin{aligned} f'(x) &= x^2 - 4x + 3 \\ &= (x - 1)(x - 3) \end{aligned}\]

  2. For the candidate \(x = 1\), test \(x = 0\) on the left and \(x = 2\) on the right, before the next cut point 3. The signs are positive, then negative.

    \[\begin{gathered} f'(0) = \cg{3} > 0 \\ f'(2) = \cr{-1} < 0 \end{gathered}\]

  3. For the candidate \(x = 3\), test \(x = 2\) on the left and \(x = 4\) on the right. The signs are negative, then positive.

    \[\begin{gathered} f'(2) = \cr{-1} < 0 \\ f'(4) = \cg{3} > 0 \end{gathered}\]

Example 2: a derivative that does not exist

Find the signs of \(f'\) on both sides of each candidate of \(f(x) = (x + 1)\sqrt[3]{x - 2}\).

  1. Use the product rule and write the result as one fraction.

    \[\begin{aligned} f'(x) &= \sqrt[3]{x - 2} + \dfrac{x + 1}{3\sqrt[3]{(x - 2)^2}} \\ &= \dfrac{3(x - 2) + (x + 1)}{3\sqrt[3]{(x - 2)^2}} \\ &= \dfrac{4x - 5}{3\sqrt[3]{(x - 2)^2}} \end{aligned}\]

  2. The top is 0 at \(x = \frac54\). The bottom is 0 at \(x = 2\), where \(f(2) = 0\) exists but \(f'(2)\) does not. Both are candidates.

    \[x = \dfrac54, \qquad x = 2\]

  3. A tempting wrong move: The derivative does not exist at \(x = 2\), so the sign of \(f'\) changes there.

    \[\begin{aligned} &f'(2)\text{ does not exist} \\ &\Rightarrow\; \text{sign change} \end{aligned}\]

    The curve is green on both sides of \(x = 2\). It rises through \((2, 0)\) with a vertical tangent, so \(f'\) is positive on both sides.

  4. The bottom is a cube root of a square, so it is positive for every \(x\) except 2. This means that \(f'\) has the sign of \(4x - 5\).

    \[\cr{-}\ \text{for}\ x < \tfrac54, \qquad \cg{+}\ \text{for}\ x > \tfrac54\]

  5. Test \(x = 0\), \(x = 1.6\) and \(x = 3\), one in each stretch. At \(\frac54\) the signs are negative, then positive. At 2 they are positive, then positive.

    \[\begin{gathered} 4(0) - 5 = \cr{-5} < 0 \\ 4(1.6) - 5 = \cg{1.4} > 0 \\ 4(3) - 5 = \cg{7} > 0 \end{gathered}\]

Practice

Find the signs of \(f'\) on both sides of each candidate of \(f(x) = x^3e^{-x}\).

The graph of \(f'\) touches the axis at 0 without crossing it. Factor \(f'\) to see why.

Solution

  1. Use the product rule and factor. The candidates are \(x = 0\) and \(x = 3\).

    \[\begin{aligned} f'(x) &= 3x^2e^{-x} - x^3e^{-x} \\ &= x^2(3 - x)e^{-x} \end{aligned}\]

  2. Except at 0, the factors \(x^2\) and \(e^{-x}\) are positive, so \(f'\) has the sign of \(3 - x\).

  3. For the candidate \(x = 0\), test \(x = -0.5\) and \(x = 1\). The signs are positive, then positive.

    \[\begin{gathered} 3 - (-0.5) = \cg{3.5} > 0 \\ 3 - 1 = \cg{2} > 0 \end{gathered}\]

  4. For the candidate \(x = 3\), test \(x = 2\) and \(x = 4\). The signs are positive, then negative.

    \[\begin{gathered} 3 - 2 = \cg{1} > 0 \\ 3 - 4 = \cr{-1} < 0 \end{gathered}\]

Questions

The upper graph shows a function \(g\) and the lower graph its derivative \(g'\). The tangent is horizontal at \(x = -2\), \(x = 1\) and \(x = 3\).

  1. What are the signs of \(g'\) just before and just after \(x = -2\), from left to right?

    • Negative, then positive
    • Positive, then negative
    • Positive, then positive
    Answer

    Negative, then positive

  2. What happens to the sign of \(g'\) at \(x = 1\), where the tangent is horizontal?

    • It is positive on both sides
    • It changes from positive to negative
    • It changes from negative to positive
    Answer

    It is positive on both sides

  3. Let \(h'(x) = \dfrac{(x - 4)(x + 1)^3}{x^2 + 1}\). What are the signs of \(h'\) just before and just after \(x = -1\)?

    • Positive, then negative
    • Negative, then positive
    • Negative, then negative
    Answer

    Positive, then negative

Summary

You choose a test point on each side of the candidate, between it and the next cut point, and find the sign of \(f'\) there.

The sign at a test point is the sign of \(f'\) on the whole side, because \(f'\) cannot change sign between cut points.

You read the two signs from left to right to see whether the curve turns at the candidate.

This is one of four skills in topic 5.4 of Which & Why: AP Calculus.