From a Candidate to an Inflection Point
Concavity describes how a graph bends. An inflection point marks a change in that bending: the graph changes from concave up to concave down, or from concave down to concave up. Finding a possible location is only part of the work. To confirm an inflection point, check the concavity on both sides, then find the corresponding function value.
The second derivative provides a way to make that comparison. Where \(f''(x)>0\) on an interval, the graph of \(f\) is concave up there. Where \(f''(x)<0\), the graph is concave down. These conclusions rely on the Second Derivative Concavity Theorem, provided the relevant derivatives exist on the interval.
A value \(c\) is a candidate when it is a location worth testing for a change in concavity. For a function whose second derivative exists and is continuous in the region being examined, zeros of \(f''\) are natural candidates. The sign check, not the candidate label by itself, determines whether concavity changes.
The Sign-Change Test
Suppose \(f\) is continuous at \(c\), and \(f''\) exists on each of two intervals immediately beside \(c\). If \(f''\) has one sign throughout the interval to the left and the opposite sign throughout the interval to the right, the Second Derivative Concavity Theorem gives opposite concavities on those intervals. Therefore the graph changes concavity at \(c\), and \((c,f(c))\) is an inflection point.
The signs have to be compared across \(c\). Finding \(f''(c)=0\) does not by itself complete that comparison. A sign chart makes the check visible: find the sign of \(f''\) on each side, translate those signs into concavity, and only then calculate the point on the graph.
Why the Test Works
The justification follows directly from the theorem connecting the second derivative and concavity. The sign of \(f''\) identifies the graph’s concavity on each side. Opposite signs therefore mean the graph has opposite concavities on those sides. Since an inflection point is where that change occurs on the graph, its coordinates must include both the input \(c\) and the output \(f(c)\).
Identify a value \(c\) where the sign of \(f''\) may change.
Determine the sign of \(f''\) on an interval just to the left and an interval just to the right of \(c\).
Positive \(f''\) means concave up; negative \(f''\) means concave down. Opposite signs give a change in concavity.
Evaluate \(f(c)\) and report \((c,f(c))\), not just \(c\).
Worked Examples
Worked Example: A Cubic with an Inflection Point at the Origin
Find and verify the inflection point of \(f(x)=x^3-3x\).
Step 1: Differentiate twice.
The candidate for a sign change is where \(f''(x)=0\):
Step 2: Check the signs on either side. Choose \(x=-1\) to test the interval to the left and \(x=1\) to test the interval to the right:
Because \(f''(x)=6x\), it is negative throughout \(x<0\) and positive throughout \(x>0\). Thus the graph changes from concave down to concave up at \(x=0\).
Step 3: Find the function value.
Conclusion and check: The inflection point is \((0,0)\). The signs \(-6\) and \(6\) confirm opposite concavities on the two sides, and the point lies on the graph because its coordinates satisfy \(y=f(0)=0\).
Worked Example: A Shifted Cubic
Find the inflection point of \(g(x)=(x-2)^3+4\), and state its \(x\)-coordinate separately from its function value.
Step 1: Differentiate twice. Applying the chain rule gives
Step 2: Locate the candidate.
Step 3: Test on both sides. Use \(x=1\) on the left and \(x=3\) on the right:
More generally, \(6(x-2)<0\) for \(x<2\) and \(6(x-2)>0\) for \(x>2\). The graph changes from concave down to concave up as \(x\) passes through \(2\).
Step 4: Evaluate the function at the candidate.
Conclusion and check: The \(x\)-coordinate is \(2\), while the function value is \(4\). The inflection point is \((2,4)\), not the number \(2\) alone. Substituting \(x=2\) into the function confirms that the graph’s point at that input has height \(4\).
Worked Example: Two Inflection Points of a Quartic
Find all inflection points of \(h(x)=x^4-6x^2\).
Step 1: Differentiate twice.
Step 2: Identify candidates. Set the factored second derivative equal to zero:
Step 3: Determine the signs in the three intervals. Test \(x=-2\), \(x=0\), and \(x=2\), one value in each interval:
Thus \(h\) is concave up on \(x<-1\), concave down on \(-1<x<1\), and concave up on \(x>1\). The concavity changes at both candidates: from up to down at \(-1\), and from down to up at \(1\).
Step 4: Calculate both function values.
Conclusion and check: The inflection points are \((-1,-5)\) and \((1,-5)\). The sign tests confirm a concavity change at each input, and substituting each input into \(h\) gives the stated output \(-5\).
Reading the Coordinates Correctly
An \(x\)-coordinate tells you where an event occurs horizontally. A function value tells you the corresponding vertical position. These are different pieces of information. If the sign test confirms a change at \(x=c\), the point on the graph is \((c,f(c))\).
| What you have found | What it means |
|---|---|
| \(c\) | The input where concavity changes |
| \(f(c)\) | The graph’s output, or height, at that input |
| \((c,f(c))\) | The inflection point on the graph |
This distinction matters in written explanations and on the AP exam. A conclusion such as “the inflection point is \(x=2\)” identifies the input but does not give the point. A complete conclusion states that the inflection point is \((2,f(2))\), and evaluates \(f(2)\) when the function is given.
Common Pitfall: Stopping at the Candidate
A common error is to find a zero of \(f''\), label it an inflection point immediately, and stop. That skips the defining check: the concavity must actually change across the candidate. Instead, determine the sign of \(f''\) on both sides, translate the signs into concavity, and evaluate the function at the confirmed input.
A second error is to report only the input. For the shifted cubic above, the incomplete conclusion “the inflection point is \(2\)” confuses a coordinate with a point. The calculation \(g(2)=4\) shows that the graph point is \((2,4)\). Showing both the sign comparison and the function value makes the conclusion clear and verifiable.
Check Your Understanding
For each problem, show the sign comparison and distinguish the candidate’s input from the point on the graph.
- For \(p(x)=x^3+6x^2\), find the candidate where \(p''(x)=0\). Test the signs of \(p''\) on either side and determine whether it is an inflection point.
- For \(q(x)=(x+1)^3-2\), determine the inflection point. State its \(x\)-coordinate and function value separately.
- For \(r(x)=x^4-2x^2\), find the zeros of \(r''\), check the signs in the intervals they determine, and identify any inflection points.
- In your own words, why does an inflection-point conclusion require comparing concavity on both sides of a candidate?
- If a sign test confirms a concavity change at \(x=c\), what additional calculation is needed to report the inflection point?
