Verify a change in concavity

Test numbers on each side

The top graph shows \(f(x) = x^4 - 6x^3\), with its tangent at the orange point. The lower graph shows \(f''(x) = 12x(x - 3)\), shaded green where \(f''\) is positive and red where it is negative. It is 0 at \(x = 0\) and at \(x = 3\), so these are the candidates, the only x-values where a point of inflection can be.

Press the first three buttons. At \(x = -1\), \(f''\) is 48, which is positive, so the top graph is concave up there. At \(x = 1\) it is \(-24\), which is negative, so the graph is concave down. At \(x = 4\) it is 48 again.

The sign of \(f''\) changes at \(x = 0\), from positive to negative, and again at \(x = 3\), from negative to positive. The concavity changes at both, so both are points of inflection. The last button fills the sign chart below and marks both.

A change of sign

The middle graph shows \(f'\), green where it rises and red where it falls. Move the point across each zero of \(f''\) and compare the shading on the two sides.

At the first and last zeros, the graph of \(f''\) crosses the axis, so the color changes. The curve changes from concave up to concave down, or the reverse.

At the middle zero, the graph of \(f''\) touches the axis and returns, so the color and the concavity stay the same. A zero of \(f''\) is only a candidate, because a point of inflection needs a change of sign.

Notation

1. Take a candidate \(c\). Choose a test number on each side of it. No other candidate may lie between a test number and \(c\), because \(f''\) could change sign there.

   \[c_{\text{left}} < c < c_{\text{right}}\]

2. Find the sign of \(f''\) at each test number and record it on the sign chart.

   \[f''(c_{\text{left}}), \qquad f''(c_{\text{right}})\]

3. If the two signs differ, the concavity changes at \(c\), and the point is a point of inflection.

   \[\begin{gathered} \cg{+} \to \cr{-} \quad\text{or}\quad \cr{-} \to \cg{+} \\ \text{inflection point} \end{gathered}\]

4. If the two signs are the same, the concavity does not change, even if \(f''(c) = 0\). Here the second zero moves onto \(c\), and \(f''\) becomes positive on both sides.

   \[\begin{gathered} \cg{+} \to \cg{+} \quad\text{or}\quad \cr{-} \to \cr{-} \\ \text{no inflection point} \end{gathered}\]

When it fails

The graph of \(f(x) = x^4 - kx^3\) has two points of inflection when \(k = 3\), at \(x = 0\) and at \(x = k/2\). Between them \(f''\) is negative. Move \(k\) toward 0.

The two zeros of \(f''\) move together, and the red stretch between them shrinks until it disappears. At \(k = 0\), \(f''(x) = 12x^2\) is 0 at \(x = 0\) and positive on both sides. The graph is concave up on both sides, so it has no point of inflection.

Example 1: A sine curve

Show that the graph of \(g(x) = \sin 2x\) has a point of inflection at \(x = \tfrac{\pi}{2}\), for \(0 < x < \pi\).

1. Differentiate twice.

   \[\begin{gathered} g'(x) = 2\cos 2x \\ g''(x) = -4\sin 2x \end{gathered}\]

2. The second derivative is 0 when \(\sin 2x = 0\). On \((0, \pi)\) this happens only at \(2x = \pi\), so \(x = \tfrac{\pi}{2}\).

3. Test one number on each side of \(\tfrac{\pi}{2}\).

   \[g''\!\left(\tfrac{\pi}{4}\right) = \cr{-4}, \qquad g''\!\left(\tfrac{3\pi}{4}\right) = \cg{4}\]

4. The sign goes from negative to positive, so the concavity changes. The graph has a point of inflection at \(x = \tfrac{\pi}{2}\).

Example 2: A zero that does not change sign

Find the points of inflection of \(f(x) = x^5 - 5x^4\).

1. Differentiate twice and factor.

   \[\begin{aligned} f'(x) &= 5x^4 - 20x^3 \\ f''(x) &= 20x^3 - 60x^2 \\ &= 20x^2(x - 3) \end{aligned}\]

2. A tempting wrong move: The second derivative is 0 at \(x = 0\) and at \(x = 3\), so the graph has a point of inflection at both.

   \[f''(0) = 0 \quad\text{and}\quad f''(3) = 0\]

   The lower graph touches the axis at \(x = 0\) and returns below it, so \(f''\) is negative on both sides. The graph is concave down on both sides, so the concavity does not change.

3. Test one number on each side of each zero.

   \[\begin{gathered} f''(-1) = \cr{-80}, \quad f''(1) = \cr{-40} \\ f''(2) = \cr{-80}, \quad f''(4) = \cg{320} \end{gathered}\]

4. The sign is negative on both sides of \(x = 0\), so the concavity does not change there. At \(x = 3\) it goes from negative to positive, so the only point of inflection is at \(x = 3\).

Practice

Show that the graph of \(k(x) = e^{-x^2/2}\) has a point of inflection at \(x = 1\) and at \(x = -1\).

Find \(k''\) and factor it. The factor \(e^{-x^2/2}\) is positive, so the shading under \(k''\) follows the other factor. Test a number on each side.

Questions

These questions use the graph of \(w(x) = \tfrac{1}{20}x^5 - \tfrac14 x^4 + 2x^2\). The top graph shows \(w\), and the lower graph shows \(w''(x) = (x + 1)(x - 2)^2\), which is 0 at \(x = -1\) and at \(x = 2\).

1. Does the graph of \(w\) have a point of inflection at \(x = -1\)?

   • Yes. The lower graph goes from below the axis to above it.
   • No. A zero of \(w''\) is not enough.
   • Yes. The lower graph is positive on both sides.
   Answer

   Yes. The lower graph goes from below the axis to above it.

2. Does the graph of \(w\) have a point of inflection at \(x = 2\)?

   • No. The lower graph touches the axis at \(x = 2\) and stays above it.
   • Yes, because \(w''(2) = 0\).
   • Yes, because the lower graph touches the axis.
   Answer

   No. The lower graph touches the axis at \(x = 2\) and stays above it.

3. The second derivative of \(f\) is \(f''(x) = x^2(x - 4)^3(x + 1)\). At which x-values does the graph of \(f\) have a point of inflection?

   • \(x = -1\) and \(x = 4\)
   • \(x = -1\), \(x = 0\) and \(x = 4\)
   • \(x = 4\) only
   Answer

   \(x = -1\) and \(x = 4\)

Summary

You choose a test number on each side of a candidate and find the sign of \(f''\) at each.

The concavity changes only if the two signs differ, so a zero of \(f''\) alone is not enough.

You do this for every candidate before you call it a point of inflection.

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