Does the graph of z have a point of inflection at x = -2?

Yes. The lower graph goes from green to red, so the sign of z'' changes.

No. The second derivative does not exist at x = -2.

No. The graph has a vertical tangent at x = -2.

Does the graph of z have a point of inflection at x = 2?

No. The lower graph is red on both sides of x = 2.

Yes, because z''(2) does not exist.

Yes, because the graph has a sharp point at x = 2.

Topic 5.6

Handle an undefined second derivative

Builds on reporting an inflection point. 6 minutes.

A bend with no second derivative

The top graph shows \(f(x) = (x - 1)^{5/3}\). The lower graph shows \(f''(x) = \dfrac{10}{9\sqrt[3]{x - 1}}\), red where \(f''\) is negative and green where it is positive.

Move the point to \(x = 0\). There \(f''(0) = -\tfrac{10}{9}\) is negative, so the top graph is concave down. At \(x = 2\), \(f''(2) = \tfrac{10}{9}\) is positive, so the top graph is concave up.

At \(x = 1\) the denominator is 0, so \(f''(1)\) does not exist, and the lower graph has no point there. The top graph is smooth at \((1, 0)\), and its tangent is flat.

The concavity changes from down to up at \(x = 1\), so the point \((1, 0)\) is a point of inflection, although \(f''(1)\) does not exist.

Two graphs, one undefined point

The numbers are gone. The graph of \(f''\) breaks at an x-value \(c\), and the open circles show that \(f''(c)\) does not exist, although \(f(c)\) is defined.

Press Case A. The shading under \(f''\) is green left of \(c\) and red right of it, so the concavity changes from up to down. This gives a point of inflection at \(c\).

Press Case B. The curve has a sharp point at \(c\), but the shading is red on both sides, so the curve is concave down on both sides. The concavity does not change, so there is no point of inflection.

The signs of \(f''\) on the two sides of \(c\) decide, not the missing value at \(c\).

Notation

Suppose that \(f''(c)\) does not exist, and that \(f\) is defined and continuous at \(c\). Then \(c\) is a candidate, an x-value where the graph may have a point of inflection.
\[\begin{gathered} \cy{f''}(c)\ \text{does not exist} \\ f \text{ continuous at } c \end{gathered}\]
The value at \(c\) does not exist, so test the sign of \(f''\) at one number on each side of \(c\).
\[f''(c_{\text{left}}), \qquad f''(c_{\text{right}})\]
If the signs differ, the concavity changes, and \((c, f(c))\) is a point of inflection, even though \(f''(c)\) does not exist.
\[\cg{+} \to \cr{-} \quad\text{or}\quad \cr{-} \to \cg{+}\]
If the signs are the same, the concavity does not change, and there is no point of inflection. Here the curve has a sharp point at \(c\) and is concave down on both sides.

When it fails

The rule needs \(f\) to be continuous at \(c\). Press the button to put a jump in \(f\) at \(c\).

The shading under \(f''\) still changes color at \(c\), so the concavity changes. But \(f\) is not continuous at \(c\), so the graph has no point of inflection there.

Example 1: A vertical tangent

Find the points of inflection of \(g(x) = 2 - \sqrt[3]{x - 3}\).

Write the cube root as a power. Then differentiate twice.
\[\begin{gathered} g'(x) = -\tfrac13 (x - 3)^{-2/3} \\ g''(x) = \tfrac29 (x - 3)^{-5/3} \end{gathered}\]
The second derivative is never 0, because its numerator is \(\tfrac29\). It does not exist at \(x = 3\), where \(g\) is continuous and \(g(3) = 2\). The only candidate is \(x = 3\).
Test one number on each side.
\[g''(2) = \cr{-\tfrac29}, \qquad g''(4) = \cg{\tfrac29}\]
The sign changes from negative to positive, so the concavity changes and \((3, 2)\) is a point of inflection, with a vertical tangent.

Example 2: A sharp point and a smooth bend

Find the points of inflection of \(h(x) = x^{5/3} + 5x^{2/3}\).

Differentiate twice and factor.
\[\begin{aligned} h'(x) &= \tfrac53 x^{2/3} + \tfrac{10}{3}x^{-1/3} \\ h''(x) &= \tfrac{10}{9}x^{-1/3} - \tfrac{10}{9}x^{-4/3} \\ &= \tfrac{10}{9}x^{-4/3}(x - 1) \end{aligned}\]
The second derivative is 0 at \(x = 1\). It does not exist at \(x = 0\), where \(h\) is continuous and \(h(0) = 0\). The candidates are \(x = 0\) and \(x = 1\).
A tempting wrong move: The second derivative does not exist at \(x = 0\), so \(h\) has a point of inflection there.
\[\begin{gathered} h''(0)\ \text{does not exist} \\ \Rightarrow\ \text{inflection at } 0 \end{gathered}\]
The lower graph is red on both sides of 0, so \(h\) is concave down on both sides of the sharp point. A sharp point is not a change of concavity.
The factor \(x^{-4/3}\) is positive for every \(x \ne 0\), so \(h''\) has the sign of \(x - 1\). Test one number in each interval of the sign chart.
\[\begin{gathered} h''(-1) = \cr{-\tfrac{20}{9}}, \quad h''\!\left(\tfrac18\right) = \cr{-\tfrac{140}{9}} \\ h''(8) = \cg{\tfrac{35}{72}} \end{gathered}\]
The sign is negative on both sides of \(x = 0\), so there is no point of inflection there. It changes from negative to positive at \(x = 1\), so the only point of inflection is \((1, 6)\).
\[h(1) = 1 + 5 = 6\]

Practice

Find the x-values where the graph of \(k(x) = x^{4/3} - 4x^{1/3}\) has a point of inflection.

The lower graph changes color at two x-values: the zero of \(k''\) and the x-value where \(k''\) does not exist. Test one number on each side of each.

Solution

The second derivative is 0 at \(x = -2\). It does not exist at \(x = 0\), where \(k\) is continuous and \(k(0) = 0\). Both are candidates.
\[\begin{gathered} k'(x) = \tfrac43 x^{1/3} - \tfrac43 x^{-2/3} \\ k''(x) = \tfrac49 x^{-2/3} + \tfrac89 x^{-5/3} \\ k''(x) = \tfrac49 x^{-5/3}(x + 2) \end{gathered}\]
Test one number in each interval of the sign chart.
\[\begin{gathered} k''(-8) = \cg{\tfrac{1}{12}}, \quad k''(-1) = \cr{-\tfrac49} \\ k''(1) = \cg{\tfrac43} \end{gathered}\]
The sign changes at \(x = -2\), and it changes again at \(x = 0\). The graph has points of inflection at \(x = -2\) and at \(x = 0\).

Questions

These questions use the graph of \(z(x) = \sqrt[3]{x + 2} + \sqrt[3]{(x - 2)^2}\). The lower graph shows \(z''\), which does not exist at \(x = -2\) or at \(x = 2\).

Does the graph of \(z\) have a point of inflection at \(x = -2\)?
- Yes. The lower graph goes from green to red, so the sign of \(z''\) changes.
- No. The second derivative does not exist at \(x = -2\).
- No. The graph has a vertical tangent at \(x = -2\).
Answer
Yes. The lower graph goes from green to red, so the sign of \(z''\) changes.
Does the graph of \(z\) have a point of inflection at \(x = 2\)?
- No. The lower graph is red on both sides of \(x = 2\).
- Yes, because \(z''(2)\) does not exist.
- Yes, because the graph has a sharp point at \(x = 2\).
Answer
No. The lower graph is red on both sides of \(x = 2\).
Which function has a point of inflection at \(x = 1\), although its second derivative does not exist there?
- \(f(x) = (x - 1)^{1/3}\)
- \(f(x) = (x - 1)^{2/3}\)
- \(f(x) = (x - 1)^{4/3}\)
- \(f(x) = (x - 1)^2\)
Answer
\(f(x) = (x - 1)^{1/3}\)

Summary

An x-value where \(f\) is continuous but \(f''\) does not exist is a candidate, and you test the sign of \(f''\) on each side.

A change of sign means a change of concavity, and the test does not need a value of \(f''\) at the point.

You use this test for graphs with cube roots, sharp points or vertical tangents, where \(f''\) can fail to exist.

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