Determine concavity on intervals

A slope that falls and then rises

The top graph shows \(f(x) = \tfrac16 x^3 - x\), with its tangent at the orange point. The lower graph shows the slope, \(f'(x) = \tfrac12 x^2 - 1\), and the height of its orange bar is the slope of that tangent. The lower curve is red where the slope falls and green where it rises.

Move the point from \(x = -2\) to \(x = 0\). The slope falls from 1 to \(-1\), so the tangent lies above the curve. The top graph is concave down there, which means it bends like a cap.

Move the point from \(x = 0\) to \(x = 2\). The slope rises from \(-1\) to 1, so the tangent lies below the curve. The top graph is concave up there, which means it bends like a cup.

At \(x = 1\) the graph falls, because the slope is \(-\tfrac12\). It is also concave up, because the slope is rising. Falling and concave up are two separate facts.

The slope decides the bend

Drag the point along this curve, which has no numbers. Where the lower curve is red, the slope falls and the tangent lies above the top graph, so the top graph is concave down.

Where it is green, the slope rises and the tangent lies below, so the top graph is concave up.

The second derivative \(f''\) is the derivative of \(f'\), so it is the slope of the lower curve. If \(f''\) is positive, the slope rises. If \(f''\) is negative, the slope falls.

Notation

1. If \(f''(x) > 0\) on an interval, then \(f'\) increases there, so the graph of \(f\) is concave up. The lowest graph is shaded green where \(f''\) is positive.

   \[\begin{gathered} \cg{f''(x) > 0} \;\Longrightarrow\; f' \text{ increasing} \\ \Longrightarrow\; \text{concave up} \end{gathered}\]

2. If \(f''(x) < 0\) on an interval, then \(f'\) decreases there, so the graph of \(f\) is concave down. The shading is red where \(f''\) is negative.

   \[\begin{gathered} \cr{f''(x) < 0} \;\Longrightarrow\; f' \text{ decreasing} \\ \Longrightarrow\; \text{concave down} \end{gathered}\]

3. To find the concavity, find \(f''\) and the x-values where \(f''\) is 0 or does not exist. They split the number line into intervals, and \(f''\) keeps one sign on each, because it can change sign only at those x-values. Test one number in each interval.

4. Whether the graph rises or falls depends on the sign of \(f'\). How it bends depends on the sign of \(f''\). The top curve is now green where it rises and red where it falls.

When it fails

Drag \(b\) to the right. At first the region under \(f''\) from \(a\) to \(b\) is all green, so the graph is concave up on the whole interval.

When \(b\) passes the first x-value where \(f''\) changes sign, the region has a green part and a red part. The graph is concave up on one and concave down on the other, so it has no single concavity on the whole interval. Split it where \(f''\) changes sign.

Example 1: Two splitting points

Find the intervals where \(f(x) = x^4 - 8x^3\) is concave up and where it is concave down.

1. Differentiate twice, and factor \(f''\).

   \[\begin{aligned} f'(x) &= 4x^3 - 24x^2 \\ f''(x) &= 12x^2 - 48x \\ &= 12x(x - 4) \end{aligned}\]

2. The second derivative is 0 at \(x = 0\) and at \(x = 4\), so these two values split the number line into three intervals.

3. Test one number in each interval.

   \[\begin{gathered} f''(-1) = \cg{60}, \quad f''(1) = \cr{-36} \\ f''(5) = \cg{60} \end{gathered}\]

4. The graph is concave up on \((-\infty, 0)\) and on \((4, \infty)\), where \(f''\) is positive, and concave down on \((0, 4)\), where \(f''\) is negative.

Example 2: A rising graph that bends down

Find the intervals where \(h(x) = 8\sqrt{x} - x\) is concave up and where it is concave down, for \(x > 0\).

1. Differentiate twice.

   \[\begin{gathered} h'(x) = \dfrac{4}{\sqrt{x}} - 1 \\ h''(x) = -\dfrac{2}{x\sqrt{x}} \end{gathered}\]

2. A tempting wrong move: On \((0, 16)\) the graph of \(h'\) is above its axis, shaded green, so \(h\) rises. A rising graph is concave up.

   \[h'(x) > 0 \;\Rightarrow\; \text{concave up}\]

   The tangent lies above the graph of \(h\). The graph of \(h'\) falls, so the slope decreases to 0 at \(x = 16\), and \(h\) is concave down.

3. Concavity depends on \(h''\), not on \(h'\). For \(x > 0\), the product \(x\sqrt{x}\) is positive, so \(h''(x)\) is negative and \(h'\) falls everywhere, shown red.

4. The graph is concave down on \((0, \infty)\). It rises on \((0, 16)\) and falls on \((16, \infty)\), and it is concave down on both.

Practice

Find the intervals where \(p(x) = \tan^{-1}x\) is concave up and where it is concave down.

The lower graph, the slope \(p'\), rises until \(x = 0\) and falls after it. Find \(p''\) and confirm where its sign changes.

Questions

The top graph shows \(u(x) = \tfrac14 x^4 - \tfrac32 x^2\), and the lower graph shows its slope, \(u'(x) = x^3 - 3x\).

1. On the interval from \(x = -1\) to \(x = 1\), the lower graph falls. What does this tell you about \(u\)?

   • The graph of \(u\) is concave down on this interval.
   • The graph of \(u\) is decreasing on this interval.
   • The graph of \(u\) is concave up on this interval.
   Answer

   The graph of \(u\) is concave down on this interval.

2. At \(x = -2\), the graph of \(u\) falls. How does it bend there?

   • It is concave up, because the lower graph rises at \(x = -2\).
   • It is concave down, because the graph falls.
   • It is concave down, because \(u'(-2)\) is negative.
   Answer

   It is concave up, because the lower graph rises at \(x = -2\).

3. On which interval is \(k(x) = \ln(1 + x^2)\) concave up?

   • \((-1, 1)\)
   • \((0, \infty)\)
   • \((-\infty, -1) \cup (1, \infty)\)
   Answer

   \((-1, 1)\)

Summary

You split the number line where \(f''\) is 0 or does not exist and test the sign of \(f''\) on each interval.

A positive \(f''\) means a rising slope and a concave up graph, and a negative \(f''\) means a falling slope and a concave down graph.

You can now find how a graph bends, which is a different question from whether it rises or falls.

This is one of five skills in topic 5.6 of Which & Why: AP Calculus.