What is an inflection point, and why is a zero second derivative not enough?

Treat it as a candidate, then check what happens to concavity on either side. Where the second derivative is positive on an interval, the graph is concave up there; where it is negative, the graph is concave down. An inflection point requires the concavity to change at the point, and the function must be continuous there. So a zero second derivative alone is not enough. [S1]

A practical test is to choose one x-value just to the left and one just to the right of the candidate, then check the sign of the second derivative at each. Opposite signs indicate a concavity change; matching signs do not. This sign test checks the intervals around the point, rather than relying on the value at the point itself.

Take f(x) = x³ and check x = 0. First differentiate: f′(x) = 3x², so f″(x) = 6x. At x = 0, f″(0) = 0, which makes 0 a candidate.

Now test either side. At x = −1, f″(−1) = −6, so the graph is concave down there. At x = 1, f″(1) = 6, so it is concave up there. The sign changes from negative to positive, so the concavity changes at 0. Since x³ is continuous there, (0, 0) is an inflection point. This is the same sign-change pattern shown for x³ in the calculus reference. [S1]

Use f(x) = x⁴ and check x = 0. Its derivatives are f′(x) = 4x³ and f″(x) = 12x², so f″(0) = 0. But for a test value on either side, such as −1 and 1, f″(−1) = 12 and f″(1) = 12. The second derivative is positive on both sides, so concavity stays up rather than changing. Therefore (0, 0) is not an inflection point. [S1]

The contrast is the key: both examples have a zero second derivative at 0, but only x³ changes the sign of f″ there. The reference supports these examples and the sign-test approach; it does not establish a rule that every candidate can be settled by checking just one pair of sample values. For a careful solution, establish the signs on the intervals around the candidate.

Write the reasoning in this order: candidate point, second derivative, sign on the left, sign on the right, and conclusion. For example:

“Candidate: x = 0. f″(x) = 6x. For x < 0 near 0, f″(x) < 0; for x > 0 near 0, f″(x) > 0. The sign changes from negative to positive, and f is continuous at 0. Therefore, x = 0 is an inflection point.”

For a non-example, replace the sign comparison with the actual result: “f″ is positive on both sides, so concavity does not change; x = 0 is not an inflection point.” If the second derivative is undefined at a point, that can also make it a candidate, but you still need to check continuity and whether concavity changes across it. [S1]