Determine derivative signs on intervals

Builds on finding where the derivative's sign can change. 6 minutes.

One test point per interval

The lower graph shows \(f'(x) = (x + 3)(x - 1)(x - 4)\), which is 0 at \(-3\), 1 and 4. These cut points, where the sign of \(f'\) can change, divide the number line into four intervals.

On each interval \(f'\) has one sign, so one test point is enough. In the first interval, test \(x = -4\), where \(f'(-4) = -40\) is negative.

In the other intervals, test \(x = 0\), \(x = 2\) and \(x = 5\), where \(f'\) is 12, \(-10\) and 32.

Each sign matches the lower graph, which lies below the axis where \(f'\) is negative. There the curve of \(f\) falls and is red. Where \(f'\) is positive the curve rises and is green.

One point speaks for the interval

The lower graph again shows \(f'\), now without numbers. The region between it and the axis is green above the axis and red below it.

Drag the orange point along one colored region. The bar grows and shrinks, but it stays on one side of the axis, because \(f'\) is never 0 and never undefined inside an interval.

The bar can change sides only at a cut point. The sign of \(f'\) at one point of an interval is therefore its sign on the whole interval.

Notation

Choose one test point \(\cy{t}\) inside an interval. Do not choose a cut point, because \(f'\) is 0 or undefined there.
Find the sign of \(f'(\cy{t})\). The sign cannot change inside the interval, so it is the sign of \(f'\) on the whole interval.
\[\begin{gathered} f'(\cy{t}) > 0 \;\Rightarrow\; \cg{+} \\ f'(\cy{t}) < 0 \;\Rightarrow\; \cr{-} \end{gathered}\]
If \(f'\) is a product, read its sign from the factors. Each factor changes sign only at its own zero, and an even number of negative factors gives a positive product.
\[(x + 3)(x - 1)(x - 4)\]
In the opening example, at \(x = 0\) the factors are positive, negative and negative, so the product is positive.
On a graph of \(f'\), the sign is the side of the axis that the graph is on. It does not matter whether the graph rises or falls.

When it fails

At the orange point, \(f'\) is positive. Suppose this one point speaks for the whole line.

The claim is wrong, because the lower graph crosses the axis at two cut points. On both sides of the middle interval the shaded region is red, so \(f'\) is negative there. Test each interval separately.

Example 1: A factor that is always positive

Find the sign of \(f'\) on each interval for \(f(x) = xe^x\).

Use the product rule to find \(f'\), and write it as a product of factors.
\[f'(x) = e^x + xe^x = (x + 1)e^x\]
An exponential is never 0 or negative, so the factor \(e^x\) is positive for every \(x\). The sign of \(f'\) is then the sign of \(x + 1\).
\[e^x > 0\]
The only cut point is \(x = -1\). For \(x < -1\) the factor \(x + 1\) is negative, so \(f'\) is negative. For \(x > -1\) it is positive, so \(f'\) is positive.
\[\cr{-}\ \text{for}\ x < -1, \qquad \cg{+}\ \text{for}\ x > -1\]

Example 2: A break in the domain

Find the sign of \(f'\) on each interval for \(f(x) = \dfrac{x^2}{x - 2}\).

Use the quotient rule to find \(f'\).
\[\begin{aligned} f'(x) &= \dfrac{2x(x - 2) - x^2}{(x - 2)^2} \\ &= \dfrac{x(x - 4)}{(x - 2)^2} \end{aligned}\]
The cut points are \(x = 0\) and \(x = 4\), where \(f'\) is 0, and \(x = 2\), where \(f\) has no value. They divide the number line into four intervals.
A tempting wrong move: The factor \(x - 2\) is negative for \(x < 2\) and positive for \(x > 2\), so the sign of \(f'\) changes at \(x = 2\).
\[\begin{gathered} (x - 2)^2 \\ \text{changes sign at } x = 2 \end{gathered}\]
The lower graph stays below the axis on both sides of \(x = 2\), so the sign does not change there. The factor is squared, so it is never negative.
On every interval \((x - 2)^2\) is positive, because it is 0 only at the cut point \(x = 2\). The sign of \(f'\) therefore comes from \(x(x - 4)\) alone.
\[(x - 2)^2 > 0\]
Multiply the signs of \(x\) and \(x - 4\). Both are negative for \(x < 0\), only \(x - 4\) is negative between 0 and 4, and both are positive for \(x > 4\).
\[\begin{gathered} \cg{+}\ (x < 0), \quad \cr{-}\ (0 < x < 2), \\ \cr{-}\ (2 < x < 4), \quad \cg{+}\ (x > 4) \end{gathered}\]

Practice

Find the sign of \(f'\) on each interval for \(f(x) = 2x - e^x\).

Set \(f'\) equal to 0 to find the cut point. Then test one point on each side and compare with the lower graph.

Solution

Find the derivative. It does not factor, so solve \(f' = 0\) to find the cut point.
\[f'(x) = 2 - e^x = 0 \;\Rightarrow\; x = \ln 2\]
As a check, \(\ln 2 \approx 0.69\). Test \(x = 0\) on the left of the cut point.
\[f'(0) = 2 - 1 = \cg{1} > 0\]
Test \(x = 1\) on the right. Because \(e\) is greater than 2, the value is negative.
\[f'(1) = 2 - e \approx \cr{-0.72} < 0\]
The sign of \(f'\) is positive for \(x < \ln 2\) and negative for \(x > \ln 2\).

Questions

The top graph shows a function \(q\) and the lower graph its derivative \(q'\). The curve of \(q\) turns at \(x = -1\) and at \(x = 2\).

On which interval is \(q'\) negative?
- \(-1 < x < 2\)
- \(x < -1\)
- \(2 < x < 5\)
Answer
\(-1 < x < 2\)
At \(x = 0\), the orange point is above the x-axis. Which statement is correct?
- \(q(0) > 0\) and \(q'(0) < 0\)
- \(q(0) > 0\) and \(q'(0) > 0\)
- \(q(0) < 0\) and \(q'(0) < 0\)
Answer
\(q(0) > 0\) and \(q'(0) < 0\)
Let \(f'(x) = (x + 4)(2 - x)e^{-x}\). On which intervals is \(f'\) negative?
- \(x < -4\) and \(x > 2\)
- \(-4 < x < 2\)
- \(x > 2\) only
Answer
\(x < -4\) and \(x > 2\)

Summary

You find the sign of \(f'\) on an interval from a test point, from the factors, or from the graph of \(f'\).

One point is enough, because \(f'\) is never 0 and never undefined inside an interval, so it cannot change sign there.

Use the signs to find where f rises and where f falls.

This is one of four skills in topic 5.3 of Which & Why: AP Calculus.