Read concavity from a derivative graph
Rising slope, bending curve
The lower graph shows \(f'(x) = 6x - x^2\), green where it rises and red where it falls. The upper graph shows \(f\).
At \(x = 1\), \(f'(1) = 5\) is above the axis, and the lower graph rises. The slope of \(f\) grows, so the tangent turns upward and the curve lies above its tangent, which means that \(f\) is concave up.
Move the point to \(x = 5\). The value \(f'(5) = 5\) is also above the axis, but the lower graph falls. The slope of \(f\) shrinks, so the tangent turns downward and the curve lies below its tangent, which means that \(f\) is concave down.
The two points have the same height on the lower graph, but it rises at one and falls at the other. The concavity changes at \(x = 3\), the peak.
The height does not matter
Drag the first slider to move the lower graph up and down. Its colors do not change, because it rises and falls in the same places.
Where the lower graph rises, \(f\) is concave up, and where it falls, \(f\) is concave down.
The concavity of \(f\) depends only on whether the graph of \(f'\) rises or falls, not on whether it lies above or below the axis.
The button colors by the sign of \(f'\) instead, which tells where \(f\) rises or falls, and those colors do change with the height.
Notation
The second derivative \(f''\) is the derivative of \(f'\), so it is the slope of the graph of \(f'\).
\[f''(x) = \cy{\text{slope of the graph of } f'}\]Where the graph of \(f'\) rises, \(f''\) is positive, so \(f\) is concave up.
\[\begin{aligned} \cg{f' \text{ rises}} \ &\Rightarrow\ f'' > 0 \\ &\Rightarrow\ f \text{ concave up} \end{aligned}\]Where the graph of \(f'\) falls, \(f''\) is negative, so \(f\) is concave down.
\[\begin{aligned} \cr{f' \text{ falls}} \ &\Rightarrow\ f'' < 0 \\ &\Rightarrow\ f \text{ concave down} \end{aligned}\]An inflection point of \(f\) is an input where the concavity changes, so on the graph of \(f'\) it is a peak or a valley.
When it fails
Flatten the peak and the valley of \(f'\) into one flat moment at \(x = 0\).
The second derivative is 0 at \(x = 0\), but the graph of \(f'\) rises on both sides, so \(f\) stays concave up. A zero of \(f''\) is an inflection point only where the graph of \(f'\) changes between rising and falling.
Example 1: a graph above the axis
The graph shows \(f'(x) = x^2 - 4x + 5\). Find where \(f\) is concave up and concave down, and its inflection point.
The graph of \(f'\) has its lowest point \((2, 1)\) where the square \((x - 2)^2\) is 0. It lies above the axis everywhere, so \(f\) increases everywhere.
\[f'(x) = (x - 2)^2 + 1\]Left of 2, the graph of \(f'\) falls and is red, so \(f\) is concave down on \((-1, 2)\).
Right of 2, the graph of \(f'\) rises and is green, so \(f\) is concave up on \((2, 5)\).
At \(x = 2\) the graph of \(f'\) changes from falling to rising, so \(f\) has an inflection point there.
Example 2: a graph below the axis
The graph shows \(f'(x) = (x - 2)(x - 7)\). Find where \(f\) is concave up.
The graph of \(f'\) is a parabola with its lowest point at \(x = 4.5\), midway between its zeros 2 and 7. It falls left of 4.5 and rises to the right.
A tempting wrong move: Between 4.5 and 7, the graph is below the axis, so \(f\) is concave down.
\[f' < 0 \ \Rightarrow\ \text{concave down}\]The slope of \(f\) grows from \(-6\) at \(x = 5\) to \(-4\) at \(x = 6\), so the tangent turns upward and the curve lies above it. A graph below the axis means that \(f\) falls, not that it is concave down.
Read the rise and fall instead: the graph rises right of 4.5, so \(f\) is concave up on \((4.5, 9)\).
\[f'(5) = -6 \qquad f'(6) = -4\]The graph falls to the left of 4.5, so \(f\) is concave down on \((0, 4.5)\).
Practice
The graph shows \(f'(x) = 2 + \sin x\) on \([0, 2\pi]\). Find the intervals where \(f\) is concave up and concave down, and the inflection points.
Move the orange bar and find where the graph changes between rising and falling.
Solution
The graph rises to its peak at \(x = \tfrac{\pi}{2}\), falls to its valley at \(x = \tfrac{3\pi}{2}\), and rises again.
\[\begin{aligned} f'(0) &= 2 & f'\!\left(\tfrac{\pi}{2}\right) &= 3 \\ f'\!\left(\tfrac{3\pi}{2}\right) &= 1 & f'(2\pi) &= 2 \end{aligned}\]The function \(f\) is concave up where the graph rises, on \(\left(0, \tfrac{\pi}{2}\right)\) and on \(\left(\tfrac{3\pi}{2}, 2\pi\right)\), and concave down where it falls, on \(\left(\tfrac{\pi}{2}, \tfrac{3\pi}{2}\right)\).
The concavity changes at \(x = \tfrac{\pi}{2}\) and \(x = \tfrac{3\pi}{2}\), so these are the inflection points.
Questions
The graph shows the derivative \(f'\) of a function \(f\), green where it rises and red where it falls.
On which interval is \(f\) concave down?
- \((-2, 2)\)
- \((0, 3.4)\), where the graph is below the axis
- \((-3.4, 0)\), where the graph is above the axis
- \((-4, -2)\) and \((2, 4)\)
Answer
\((-2, 2)\)
How many inflection points does \(f\) have?
- 2
- 3
- 1
- 0
Answer
2
The derivative of \(f\) is \(f'(x) = x^3 - 6x^2\). On which interval is \(f\) concave down?
- \((0, 4)\)
- \((-\infty, 6)\)
- \((-\infty, 0)\) and \((4, \infty)\)
- \((0, 6)\)
Answer
\((0, 4)\)
Summary
You read where the graph of \(f'\) rises and falls: \(f\) is concave up where it rises and concave down where it falls.
This works because \(f''\) is the slope of the graph of \(f'\), so the height above or below the axis does not matter.
You use this when a problem gives the graph of \(f'\) and asks about the concavity of \(f\).
This is one of five skills in topic 5.8 of Which & Why: AP Calculus.