Why a Second-Derivative Sign Change Matters
A graph’s concavity describes how its shape bends. Where the graph is concave up, its slopes increase as you move from left to right; where it is concave down, its slopes decrease. An inflection point is a point on the graph where that concavity changes.
The second derivative is a useful way to track this change. Where \(f''(x)>0\), the function is concave up; where \(f''(x)<0\), it is concave down. But finding a zero of \(f''\), or a point where \(f''\) is undefined, only identifies a possible location. To establish an inflection point, check what happens to concavity on both sides.
Finding and Testing Candidates
For the usual AP problems involving functions with familiar derivatives, first find the \(x\)-values where \(f''(x)=0\) and where \(f''\) is undefined. These are candidates to test. Then determine the sign of \(f''\) on intervals immediately to the left and right of each candidate. A sign change establishes a change in concavity; the same sign on both sides means there is no inflection point there.
For example, a change from positive to negative means the graph changes from concave up to concave down. A change from negative to positive means it changes from concave down to concave up. The \(y\)-coordinate of a confirmed inflection point is \(f(c)\), so report the point as \((c,f(c))\), not merely as the \(x\)-value \(c\).
Solve \(f''(x)=0\) and identify where \(f''\) is undefined in the domain.
Use the candidates to divide the domain into intervals on which the sign of \(f''\) can be tested.
Determine whether \(f''\) is positive or negative on each adjacent interval.
If concavity changes and \(f\) is continuous at the candidate \(c\), calculate \(f(c)\). Otherwise, do not report \((c,f(c))\) as an inflection point.
Why the Sign Test Works
The second derivative describes how the first derivative changes. If \(f''(x)>0\) throughout an interval, then \(f'\) is increasing there. To see why, choose any \(x_1<x_2\) in that interval. The Mean Value Theorem applied to \(f'\) gives a number \(\xi\) between \(x_1\) and \(x_2\) such that
Both factors on the right are positive, so \(f'(x_2)>f'(x_1)\). Thus the slopes of \(f\) increase from left to right, which is the concave-up case. If \(f''<0\) throughout the interval, the same equation gives \(f'(x_2)<f'(x_1)\), so the slopes decrease and \(f\) is concave down.
Consequently, when the sign of \(f''\) changes across a candidate, the graph changes concavity. Checking both sides matters: the value \(f''(c)\), even if it exists, does not by itself describe what happens throughout the neighboring intervals.
Worked Example: A Sign Change at Two Candidates
Worked Example: A Sign Change at Two Candidates
Find and verify the inflection points of \(f(x)=x^4-4x^2\).
Differentiate twice:
Set the second derivative equal to zero:
The second derivative is defined for every real \(x\), so these are the only candidates. Since \(f''(x)=4(3x^2-2)\), it is positive when \(x^2>\frac{2}{3}\) and negative when \(x^2<\frac{2}{3}\). Thus it is positive on \((-\infty,-\sqrt{2/3})\), negative on \((-\sqrt{2/3},\sqrt{2/3})\), and positive on \((\sqrt{2/3},\infty)\). The sign changes at both candidates, so both are inflection points.
To find their \(y\)-coordinates, use \(x^2=\frac{2}{3}\) and \(x^4=\frac{4}{9}\):
The points are \(\left(-\sqrt{2/3},-20/9\right)\) and \(\left(\sqrt{2/3},-20/9\right)\). As a check, substituting either candidate into \(12x^2-8\) gives \(12(2/3)-8=8-8=0\), and the sign chart confirms a change at each one.
Worked Example: A Zero Without an Inflection Point
Worked Example: A Zero Without an Inflection Point
Determine whether \(x=0\) is an inflection point of \(f(x)=x^4\).
Differentiate twice:
The only candidate is \(x=0\), where \(f''(0)=0\). For every \(x<0\) and every \(x>0\), \(12x^2>0\). The graph is concave up on both sides, so concavity does not change at zero. Therefore, \(x=0\) is not an inflection point.
The point on the graph is \((0,f(0))=(0,0)\), but being on the graph does not make it an inflection point. As a check, \(f''(-1)=12>0\) and \(f''(1)=12>0\); the signs on opposite sides agree.
Worked Example: An Inflection Point Where the Second Derivative Is Undefined
Worked Example: An Inflection Point Where the Second Derivative Is Undefined
Determine whether \(x=0\) is an inflection point of \(f(x)=x^{5/3}\), defined for all real \(x\).
For \(x\ne0\), differentiate:
The second derivative is undefined at zero, so zero is a candidate. For \(x<0\), \(x^{-1/3}<0\), giving \(f''(x)<0\). For \(x>0\), \(x^{-1/3}>0\), giving \(f''(x)>0\). Thus the graph changes from concave down to concave up across zero. Since \(f(0)=0\) and \(f\) is continuous there, \((0,0)\) is an inflection point.
As a check, \(f''(-1)=-10/9\) and \(f''(1)=10/9\), confirming opposite signs on either side. The fact that \(f''(0)\) is undefined does not prevent an inflection point; the one-sided concavity test establishes the change.
Common Pitfalls and Careful Interpretation
A frequent error is to stop after solving \(f''(x)=0\). For \(x^4\), that gives \(x=0\), but \(f''\) is positive on both sides, so there is no change in concavity. The correction is to test signs on intervals, not just substitute the candidate into the second derivative.
A second error is to assume that every point where \(f''\) is undefined is an inflection point. For example, \(f(x)=1/x\) has \(f''(x)=2/x^3\), which is undefined at zero, and \(f''\) has opposite signs on the two sides. But \(f\) itself is not defined at zero, so there is no graph point \((0,f(0))\) and no inflection point there. Check that the original function is defined and continuous at a candidate before reporting it as an inflection point.
For standard AP problems, solving \(f''=0\) and identifying where \(f''\) is undefined is the usual way to find candidates. The sign chart is what verifies them. If a problem supplies a graph or sign information directly, use that information to determine concavity rather than assuming that a candidate must satisfy \(f''=0\).
Why This Analysis Matters
Inflection points describe a change in the behavior of slopes, not necessarily a maximum or minimum of the function. They help organize a graph analysis: a sign chart for \(f''\) shows where the graph bends upward or downward, while confirmed sign changes identify where that bending switches. In a written response, state the evidence explicitly: give the candidate, report the signs on its two sides, and conclude whether concavity changes.
A complete conclusion distinguishes the \(x\)-value from the point on the graph. For example, one might write: “Since \(f''\) changes from negative to positive at \(x=c\), the graph changes from concave down to concave up. Because \(f\) is continuous there, \((c,f(c))\) is an inflection point.” If the signs do not change, say that the candidate is not an inflection point.
Check Your Understanding
For each question, use concavity on both sides rather than relying only on the value of the second derivative at a candidate.
- For \(f(x)=x^3\), identify the candidate at \(x=0\) and determine whether it is an inflection point.
- For \(f(x)=x^4\), why does \(f''(0)=0\) fail to establish an inflection point?
- If \(f''\) is negative immediately to the left of \(c\) and positive immediately to the right, what change in concavity occurs?
- Can a point where \(f''\) is undefined be an inflection point? State the other conditions you would check.
- Why is \(x=0\) not an inflection point of \(f(x)=1/x\), even though \(f''\) has opposite signs on the two sides?
