What Makes a Point an Inflection Point?
A graph can change from bending upward to bending downward, or from bending downward to bending upward. A point where this change in concavity occurs is an inflection point. To justify one, it is not enough to find a place where the second derivative is zero. You must establish that concavity actually changes across the point.
The second derivative is useful because its sign describes concavity on intervals where the function is twice differentiable. The key test is therefore a comparison: determine the sign of \(f''\) just to the left of a candidate and just to the right. Opposite signs show a change in concavity, provided the function is continuous at the point.
From the Second Derivative to Concavity
On an interval where \(f\) is twice differentiable, a positive second derivative means the first derivative is increasing, so the graph is concave up. A negative second derivative means the first derivative is decreasing, so the graph is concave down.
Here is why the test works. Choose any two numbers \(u\) and \(v\) in an interval where \(f''>0\), with \(u<v\). The Mean Value Theorem applied to \(f'\) gives some \(\xi\) between \(u\) and \(v\) such that
Both factors on the right are positive: \(f''(\xi)>0\) by assumption, and \(v-u>0\). Therefore \(f'(v)>f'(u)\). Since this holds for every \(u<v\), the derivative is increasing throughout the interval, which is the derivative criterion for concavity up. If \(f''<0\), the same equation gives \(f'(v)<f'(u)\), so the derivative is decreasing and the graph is concave down.
This reasoning applies separately on each side of a possible inflection point. If the second derivative is negative on the left and positive on the right, the graph changes from concave down to concave up. If the signs are positive on the left and negative on the right, the graph changes from concave up to concave down.
Finding and Testing Candidates
A common AP procedure is to find interior points in the domain where \(f''(x)=0\) or \(f''(x)\) does not exist, and then test concavity on either side. These are candidates, not automatic inflection points. A candidate must pass the sign-change test, and the function must be defined and continuous at the point.
To make the comparison, divide the domain into intervals at the candidates. Determine the sign of \(f''\) on each interval. A sign chart can help organize the evidence, but its intervals must be chosen correctly: zeros and points where \(f''\) is undefined can separate intervals with different signs. Testing one value in an interval is useful when the expression and its domain ensure the sign stays the same throughout that interval.
Find interior domain points where \(f''=0\) or \(f''\) is undefined. Keep only points where \(f\) itself is defined.
Confirm that \(f\) is continuous at each point you may identify as an inflection point.
Determine the sign of \(f''\) on intervals immediately to the left and right of the candidate.
Opposite signs establish a change in concavity and, with continuity, an inflection point. Matching signs do not.
Under the usual AP assumptions that \(f''\) is continuous on the intervals being tested, a sign change in \(f''\) must pass through zero or a point where the second derivative is undefined. If those regularity assumptions are not available, the candidate procedure alone is not a universal theorem; concavity must be examined directly.
Worked Example: A Concavity Change at a Zero of the Second Derivative
Worked Example: A Concavity Change at a Zero of the Second Derivative
Determine whether \(f(x)=x^3-3x\) has an inflection point at \(x=0\).
First differentiate:
The second derivative is zero when \(6x=0\), so \(x=0\) is a candidate. The function is a polynomial, so it is continuous at \(0\). For \(x<0\), \(6x<0\), hence \(f''(x)<0\) and the graph is concave down. For \(x>0\), \(6x>0\), hence \(f''(x)>0\) and the graph is concave up.
The second-derivative sign changes from negative to positive, so concavity changes from down to up. The point on the graph is
Therefore, \((0,0)\) is an inflection point. The function value confirms the point’s coordinates, and the opposite signs of \(f''\) on the two sides provide the concavity justification.
A Zero of the Second Derivative Is Not Enough
The equation \(f''(c)=0\) identifies a possible location for a concavity change. It does not prove that one occurs. If \(f''\) has the same sign on both sides, the graph remains concave in the same direction.
Worked Example: A Candidate That Is Not an Inflection Point
Determine whether \(f(x)=x^4\) has an inflection point at \(x=0\).
Differentiate twice:
Since \(12x^2=0\) at \(x=0\), zero is a candidate. The function is continuous there, with
For every \(x<0\), \(x^2>0\), so \(f''(x)=12x^2>0\). For every \(x>0\), \(x^2>0\) as well, so \(f''(x)=12x^2>0\). Thus the graph is concave up on both sides of zero. Concavity does not change, so \((0,0)\) is not an inflection point.
The result is checked directly from the signs: neither side has a negative second derivative. The zero of \(f''\) marks a candidate only; it does not establish an inflection point.
When the Second Derivative Is Undefined
A candidate can also occur where \(f''\) is undefined, as long as the function itself is defined and continuous there. The sign comparison still works if \(f''\) exists on each side. In that situation, do not substitute the candidate into a formula that is undefined; instead, analyze the signs in the two neighboring intervals.
Worked Example: An Inflection Point Where the Second Derivative Is Undefined
Determine whether \(f(x)=\sqrt[3]{x}\) has an inflection point at \(x=0\).
The cube-root function is defined and continuous at zero, and \(f(0)=0\). For \(x\ne 0\), write \(f(x)=x^{1/3}\) and differentiate:
The formula for \(f''\) is undefined at zero, so zero is a candidate. If \(x<0\), then \(x^{5/3}<0\), so \(x^{-5/3}<0\). Therefore
If \(x>0\), then \(x^{-5/3}>0\), and consequently
The graph is concave up to the left of zero and concave down to the right. Since the function is continuous at zero and \(f(0)=0\), \((0,0)\) is an inflection point. The undefined second derivative at the point does not prevent this conclusion; the sign change on its two sides establishes it.
Common Pitfall: Treating a Candidate as a Conclusion
A frequent error is to stop as soon as a zero or undefined value of \(f''\) is found. For example, a student might write, “For \(f(x)=x^4\), \(f''(0)=0\), so \((0,0)\) is an inflection point.” The first statement is correct, but the conclusion does not follow. The calculation only identifies a candidate.
The correction is to compare signs on both sides: for \(x<0\), \(12x^2>0\), and for \(x>0\), \(12x^2>0\). Both intervals are concave up, so there is no change in concavity and no inflection point at zero.
Another incomplete justification says only that “\(f''\) changes sign.” A complete response identifies the signs on the two intervals and connects them to concavity. For example: “Because \(f''<0\) immediately to the left of \(c\), the function is concave down there; because \(f''>0\) immediately to the right, it is concave up there. The function is continuous at \(c\), so \((c,f(c))\) is an inflection point.”
Writing a Complete Justification
A strong AP explanation includes evidence, not just a label. It names the candidate, gives the relevant signs of \(f''\) on both sides, identifies the corresponding concavity, and checks continuity at the point. If the candidate is an inflection point, report the coordinate \((c,f(c))\), not only the \(x\)-value.
For a candidate that fails the test, state that the signs do not change and identify the concavity on both sides. This makes clear why the candidate is not an inflection point, rather than merely reporting that a test failed.
Check Your Understanding
For each question, focus on the evidence on both sides of the proposed point.
- For \(f(x)=x^3\), find the candidate at which \(f''(x)=0\), determine the signs of \(f''\) on either side, and state whether the graph has an inflection point there.
- For \(g(x)=x^4\), explain why \(g''(0)=0\) does not make \((0,0)\) an inflection point.
- What two checks are needed, besides identifying a candidate, to justify an inflection point?
- Can an inflection point occur where the second derivative is undefined? Give the condition that still must hold at the point.
- What concavity change corresponds to a second derivative that is positive on the left and negative on the right?
