Classify a candidate using the sign change

From a sign change to a verdict

The graph shows \(f(x) = x^5 - 5x\), green where the curve rises and red where it falls. The derivative \(f'(x) = 5(x^4 - 1)\) is 0 at \(x = -1\) and at \(x = 1\), so these are the candidates. Press the first button. The line under the lower graph then shows the sign of \(f'\) on each side of \(x = -1\) in the same colors.

At \(x = -1\), \(f'\) changes from positive to negative, so the curve rises and then falls. The point \((-1, 4)\) is higher than the points near it, so \(f\) has a local maximum there.

Press the second button. At \(x = 1\), \(f'\) changes from negative to positive, so the curve falls and then rises. The point \((1, -4)\) is lower than the points near it, so \(f\) has a local minimum there.

Two signs, one verdict

The scene is the same without numbers. The orange point marks a candidate \(c\), and the line under the lower graph shows the sign of \(f'\) on each side of \(c\).

Press the first two buttons. When the signs change, the curve turns at c and forms a top or a bottom.

Press the last two buttons. When the signs are the same, the curve continues in one direction through c, so the point is neither a maximum nor a minimum.

Notation

1. If \(f'\) changes from positive to negative at \(c\), the curve rises and then falls, so \(f\) has a local maximum at \(c\).

   \[\begin{gathered} \cg{+}\ \text{then}\ \cr{-} \\ \Rightarrow\ \text{local maximum at}\ c \end{gathered}\]

2. If \(f'\) changes from negative to positive at \(c\), the curve falls and then rises, so \(f\) has a local minimum at \(c\).

   \[\begin{gathered} \cr{-}\ \text{then}\ \cg{+} \\ \Rightarrow\ \text{local minimum at}\ c \end{gathered}\]

3. If \(f'\) has the same sign on both sides of \(c\), the curve continues in one direction. Then \(f\) has neither a maximum nor a minimum at \(c\).

   \[\begin{gathered} \text{same sign on both sides} \\ \Rightarrow\ \text{neither} \end{gathered}\]

4. These rules need \(f\) to have a value at \(c\). They also work where \(f'(c)\) does not exist, as at a sharp point.

When it fails

Press the button to draw \(f(x) = 1/x^2\). Its derivative \(f'(x) = -2/x^3\) changes from positive to negative at \(x = 0\), so the rule would give a local maximum.

But \(f\) has no value at 0, and the curve rises without bound on both sides, so there is no highest point.

Example 1: one candidate

Classify the candidate of \(f(x) = \dfrac{\ln x}{x}\), for \(x > 0\).

1. Use the quotient rule to find the derivative.

   \[f'(x) = \dfrac{1 - \ln x}{x^2}\]

2. The bottom is positive. The top is 0 when \(\ln x = 1\), so the candidate is \(x = e\).

   \[1 - \ln x = 0 \;\Rightarrow\; x = e\]

3. Test \(x = 1\) on the left and \(x = e^2\) on the right.

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

4. Positive then negative means that the curve rises and then falls, so \(f\) has a local maximum at \(x = e\).

Example 2: two candidates

Classify each candidate of \(f(x) = (x - 3)^2e^x\).

1. Use the product rule and factor. The factor \(e^x\) is positive, so \(f'\) is 0 only at \(x = 1\) and \(x = 3\), the candidates.

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

2. Test \(x = 0\), \(x = 2\) and \(x = 3.5\), one before, one between and one after the candidates.

   \[\begin{gathered} f'(0) = \cg{3} > 0 \\ f'(2) = \cr{-e^2} < 0 \\ f'(3.5) = \cg{\tfrac54e^{3.5}} > 0 \end{gathered}\]

3. A tempting wrong move: At \(x = 1\) the sign goes from positive to negative. The derivative goes down, so \(f\) has a local minimum.

   \[\begin{gathered} \cg{+}\ \text{then}\ \cr{-} \\ \Rightarrow\ \text{local minimum} \end{gathered}\]

   The derivative goes down, but the curve does not: positive then negative means that the curve rises and then falls. The graph shows a top at \(x = 1\), which is a local maximum.

4. At \(x = 1\) the signs are positive, then negative, so \(f\) has a local maximum there.

5. At \(x = 3\) the signs are negative, then positive, so \(f\) has a local minimum there.

Practice

Classify each candidate of \(f(x) = x^{5/3} - 5x^{2/3}\).

A cube root of a negative number is negative. Use this for the sign of \(f'\) to the left of 0.

Questions

The graph shows the derivative \(k'\) of a function \(k\). It crosses the x-axis at \(x = -1\) and \(x = 4\). At \(x = 2\) it touches the axis from below and stays below it.

1. What does \(k\) have at \(x = -1\)?

   • A local maximum
   • A local minimum
   • Neither
   Answer

   A local maximum

2. What does \(k\) have at \(x = 2\)?

   • Neither a local maximum nor a local minimum
   • A local maximum
   • A local minimum
   Answer

   Neither a local maximum nor a local minimum

3. Let \(m'(x) = (3 - x)(x^2 + 1)\). What does \(m\) have at \(x = 3\)?

   • A local maximum
   • A local minimum
   • Neither
   Answer

   A local maximum

Summary

You read the signs of f′ on the two sides of a candidate. Positive then negative gives a local maximum, negative then positive gives a local minimum, and the same sign gives neither.

A positive slope means that the curve rises, and a negative slope means that it falls. The order of the two signs then makes a top, a bottom or neither.

You use this at every candidate, including one where \(f'\) does not exist, as long as \(f\) has a value there.

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