Find candidate inflection locations
Where the bending changes
The top graph shows \(f(x) = x^3 - 9x^2 + 12x\), with its tangent at the orange point. The lower graph shows the second derivative \(f''(x) = 6x - 18\), and its orange bar has height \(f''(x)\). The shaded region is red where \(f''\) is negative and green where it is positive.
Drag the point from the left. While \(f''\) is negative, the top graph is concave down. While \(f''\) is positive, it is concave up.
At \(x = 3\) the orange bar has length 0, because \(f''(3) = 0\). The lower line crosses the axis there, and the top graph changes from concave down to concave up.
A point where the concavity changes is a point of inflection, so this graph has one at \(x = 3\), where \(f''\) is 0.
Where the second derivative changes sign
The numbers are gone, and the middle graph now shows \(f'\), green where it rises and red where it falls. The shading under \(f''\) is green where \(f'' > 0\) and red where \(f'' < 0\). Move the point along the curve.
The concavity of the curve changes at three x-values. At two of them, the graph of \(f''\) crosses the axis, so \(f'' = 0\).
At the third, \(f'\) has a corner, so the graph of \(f''\) jumps across the axis and \(f''\) does not exist there. Where \(f''\) exists and is never 0, it keeps one sign. The concavity can change only at a zero of \(f''\) or where \(f''\) does not exist.
Notation
A point of inflection is a point on the graph where \(f\) is continuous and the concavity changes. The concavity changes at three x-values here.
\[\begin{gathered} \bigl(c,\ f(c)\bigr) \ \text{ with the concavity} \\ \text{of } f \text{ changing at } c \end{gathered}\]A candidate is an x-value \(c\) in the domain of \(f\) where \(f''(c) = 0\) or \(f''(c)\) does not exist. The sign chart marks the three candidates here.
\[\begin{gathered} \cy{f''}(c) = 0 \quad\text{or} \\ \cy{f''}(c)\ \text{does not exist} \end{gathered}\]To find the candidates, find \(f''\) and solve \(f''(x) = 0\). Then find where \(f''(x)\) is undefined, for example where a denominator is 0 or \(f'\) has a corner.
Keep only the x-values where \(f\) itself is defined. A candidate is not yet an inflection point, because you still need to test the sign of \(f''\) on both sides.
When it fails
The concavity of this curve changes at the first x-value where \(f''\) is 0. Press the button to remove the point of the graph at that x-value.
The open ring shows that \(f\) is now undefined there. The concavity still changes, but with no point on the graph there is no point of inflection. An x-value outside the domain of \(f\) is not a candidate.
Example 1: A quartic
Find the candidates for points of inflection of \(f(x) = x^4 - 18x^2\).
Differentiate twice. The second derivative is a polynomial, so it exists for every \(x\).
\[\begin{gathered} f'(x) = 4x^3 - 36x \\ f''(x) = 12x^2 - 36 \end{gathered}\]The only candidates are the zeros of \(f''\), so solve \(f''(x) = 0\).
\[\begin{gathered} 12(x^2 - 3) = 0 \\ x = -\sqrt{3} \ \text{ or } \ x = \sqrt{3} \end{gathered}\]The function \(f\) is defined at both numbers, so the candidates are \(x = -\sqrt{3}\) and \(x = \sqrt{3}\).
Example 2: A second derivative that does not exist
Find the candidates for points of inflection of \(g(x) = (x - 2)\,|x - 2|\).
Write \(g\) without the absolute value, and differentiate twice on each side of \(x = 2\).
\[\begin{gathered} g(x) = \begin{cases} (x - 2)^2 & x \ge 2 \\ -(x - 2)^2 & x < 2 \end{cases} \\ g''(x) = \begin{cases} 2 & x > 2 \\ -2 & x < 2 \end{cases} \end{gathered}\]A tempting wrong move: The second derivative is 2 or \(-2\), never 0, so \(g\) has no candidates.
\[g''(x) \ne 0 \ \text{ for every } x\]The shading under \(g''\) turns from red to green at \(x = 2\), and the tangent moves from above the curve to below it. The concavity changes at \(x = 2\), so a candidate is missing.
Left of 2, \(g''\) is \(-2\), and right of 2 it is 2. The two sides differ, so \(g''(2)\) does not exist. The function \(g\) is defined at 2, so \(x = 2\) is a candidate.
\[g(2) = 0\]
Practice
Find the candidates for points of inflection of \(f(x) = x^2 \ln x\), for \(x > 0\).
Differentiate twice and solve \(f''(x) = 0\). The lower graph crosses the axis once, so expect one candidate.
Solution
Differentiate twice. The second derivative exists for every \(x > 0\), which is the whole domain.
\[\begin{gathered} f'(x) = 2x\ln x + x \\ f''(x) = 2\ln x + 3 \end{gathered}\]Set \(f''(x) = 0\) and solve for \(x\).
\[\ln x = -\tfrac32 \quad\Rightarrow\quad x = e^{-3/2}\]This number is positive, so it is in the domain and is the only candidate. As a check, \(e^{-3/2} \approx 0.223\).
Questions
These questions use \(v(x) = x^4 - 2x^3 - 12x^2\). The top graph shows \(v\), and the lower graph shows \(v''\).
Which x-values are candidates for points of inflection of \(v\)?
- \(x = -1\) and \(x = 2\)
- Only \(x = 0\)
- \(x = 0\) and the two other x-values where the top graph has a flat tangent
Answer
\(x = -1\) and \(x = 2\)
The top graph has a hilltop at \(x = 0\). Is \(x = 0\) a candidate for a point of inflection?
- No. The lower graph is below the axis at \(x = 0\), so \(v''(0) \ne 0\).
- Yes, because the tangent at \(x = 0\) is flat.
- Yes, because \(v\) changes from rising to falling at \(x = 0\).
Answer
No. The lower graph is below the axis at \(x = 0\), so \(v''(0) \ne 0\).
How many candidates for points of inflection does \(m(x) = \dfrac{x^2}{x - 3}\) have?
- None
- One, at \(x = 3\)
- Two, at \(x = 0\) and \(x = 6\)
Answer
None
Summary
You find \(f''\) and list the x-values in the domain where \(f''\) is 0 or does not exist.
The second derivative can change sign only at a zero or where it does not exist, so the concavity can change only at these x-values.
You make this list first when a question asks where the graph can have a point of inflection.
This is one of five skills in topic 5.6 of Which & Why: AP Calculus.