Read function direction from a derivative graph
Which way does f go?
The upper graph shows \(f\), green where it rises and red where it falls. The lower graph shows its derivative \(f'\), with the shaded region green above the axis and red below it.
Move the point to \(x = 0\). The lower graph is above the axis, at height 1.5, so the slope of \(f\) is 1.5 and \(f\) rises.
At \(x = 2\), the lower graph is below the axis, at height \(-1\), so the slope of \(f\) is \(-1\) and \(f\) falls.
At \(x = 1\), the lower graph crosses the axis from above to below. The slope of \(f\) changes from positive to negative, so \(f\) has a peak.
The height of the lower graph is a slope, not a height of \(f\).
Above the axis, f rises
Drag the point. Where the graph of \(f'\) is above the x-axis, \(f'\) is positive and \(f\) rises. Where it is below, \(f'\) is negative and \(f\) falls.
At a crossing, \(f'\) changes sign, so \(f\) changes direction. When \(f'\) crosses from above to below, the curve \(f\) has a peak. When \(f'\) crosses from below to above, it has a valley.
Notation
On an interval where \(f'\) is positive, \(f\) increases. On an interval where \(f'\) is negative, \(f\) decreases.
\[\begin{aligned} \cg{f' > 0} &\Rightarrow\ f \text{ increases} \\ \cr{f' < 0} &\Rightarrow\ f \text{ decreases} \end{aligned}\]Mark the inputs where the graph of \(f'\) meets the x-axis. They split the axis into intervals. On each one the graph stays above or below the axis, because \(f'\) can change sign only where it is 0.
A turning point of \(f\) is an input where \(f\) changes between increasing and decreasing. A local maximum is a peak, where \(f'\) changes from positive to negative. A local minimum is a valley, where it changes from negative to positive.
\[\begin{aligned} + \to - &:\ \text{local maximum} \\ - \to + &:\ \text{local minimum} \end{aligned}\]
When it fails
A zero of \(f'\) gives a turning point only if \(f'\) changes sign there. Squeeze the two crossings at 0 and 2 until they meet.
At \(x = 1\), the graph of \(f'\) touches the axis and stays above it. The curve \(f\) rises on both sides of 1, and its tangent is horizontal at that one input only.
Example 1: a parabola above the axis
The graph shows \(f'(x) = 4 - x^2\). Find where \(f\) increases and decreases, and where it has turning points.
The graph of \(f'\) crosses the x-axis at \(x = -2\) and \(x = 2\), because \(4 - x^2 = 0\) there. These two inputs split the axis into three intervals.
\[4 - x^2 = 0 \ \Rightarrow\ x = -2 \ \text{ or }\ x = 2\]Between \(-2\) and 2, the graph is above the axis, so \(f\) increases. The shaded region and the curve \(f\) are green there.
Left of \(-2\) and right of 2, the graph is below the axis, so \(f\) decreases. Both are red there.
At \(x = -2\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(x = 2\), \(f'\) changes from positive to negative, so \(f\) has a local maximum.
Example 2: a graph that touches the axis
The graph shows \(f'(x) = \dfrac{(x + 1)^2 (x - 2)}{4}\). Find where \(f\) has turning points.
The graph of \(f'\) meets the x-axis at \(x = -1\) and at \(x = 2\), so \(f'\) is 0 at both inputs.
\[f'(-1) = 0 \qquad f'(2) = 0\]A tempting wrong move: The derivative is 0 at both inputs, so \(f\) has a turning point at both.
\[\begin{aligned} f'(-1) &= 0 \\ f'(2) &= 0 \\ &\Rightarrow\ \text{two turning points} \end{aligned}\]At \(x = -1\), the graph touches the axis and returns below it, so \(f'\) is negative on both sides. The function \(f\) falls on both sides and does not turn.
Test one input in each interval. The sign of \(f'\) at that input is its sign on the whole interval, because \(f'\) is 0 only at \(-1\) and 2.
\[\begin{aligned} f'(-2) &= -1 \\ f'(0) &= -\tfrac{1}{2} \\ f'(3) &= 4 \end{aligned}\]The sign is negative on both sides of \(-1\), so \(f\) does not turn there. At 2, the sign changes from negative to positive, so \(f\) has one local minimum at \(x = 2\).
Practice
The graph shows \(f'(x) = \sin x - \dfrac{1}{2}\) on \([0, 2\pi]\). Find where \(f\) increases and decreases, and classify its turning points.
Find where the graph crosses the axis. Then find the sign of the graph on each interval.
Solution
The graph crosses the axis where \(\sin x = \tfrac12\). This happens at two inputs.
\[x = \frac{\pi}{6} \qquad x = \frac{5\pi}{6}\]Test one input in each interval. The derivative is negative at 0, positive at \(\tfrac{\pi}{2}\), and negative at \(\pi\).
\[\begin{aligned} f'(0) &= -\tfrac{1}{2} \\ f'\!\left(\tfrac{\pi}{2}\right) &= \tfrac{1}{2} \\ f'(\pi) &= -\tfrac{1}{2} \end{aligned}\]The function \(f\) decreases on \(\left[0, \tfrac{\pi}{6}\right]\), increases on \(\left[\tfrac{\pi}{6}, \tfrac{5\pi}{6}\right]\), and decreases on \(\left[\tfrac{5\pi}{6}, 2\pi\right]\).
At \(\tfrac{\pi}{6}\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(\tfrac{5\pi}{6}\), it changes from positive to negative, so \(f\) has a local maximum.
Questions
The graph shows the derivative \(f'\) of a function \(f\). It touches the x-axis at \(x = 1\) and crosses the axis at \(x = -2\) and \(x = 3\).
On which interval is \(f\) increasing?
- \((-2, 3)\)
- \((-2, 1)\)
- \((1, 3)\)
- \((-\infty, -2)\)
Answer
\((-2, 3)\)
How many turning points does \(f\) have?
- 2
- 3
- 1
Answer
2
The derivative of \(f\) is \(f'(x) = x^2 - 4x - 5\). On which interval is \(f\) decreasing?
- \((-1, 5)\)
- \((-\infty, -1)\) and \((5, \infty)\)
- \((-5, 1)\)
- \((-\infty, 5)\)
Answer
\((-1, 5)\)
Summary
Above the axis \(f\) increases and below it \(f\) decreases, because the height of \(f'\) is the slope of \(f\).
Where the graph of \(f'\) crosses the axis, the slope changes sign, so \(f\) has a turning point.
You use this when a problem gives the graph of a derivative and asks about the function.
This is one of five skills in topic 5.8 of Which & Why: AP Calculus.