The second derivative test fails when the second derivative equals zero at a critical point. It also fails when the second derivative doesn’t exist at that point. In either case, the test is called “inconclusive,” meaning it cannot tell you whether the critical point is a local maximum, a local minimum, or neither.
The Exact Condition for Failure
Recall how the test normally works: you find a critical point where f'(c) = 0, then plug that point into the second derivative. If f”(c) is positive, you have a local minimum. If f”(c) is negative, you have a local maximum. The test fails when f”(c) = 0, because zero curvature at a single point doesn’t reveal whether the graph is curving up, curving down, or flattening out entirely.
The tricky part is that when f”(c) = 0, anything can happen. The critical point might still be a local max, a local min, or just an inflection point where the curve changes concavity. You genuinely cannot tell from the second derivative alone. This is why textbooks use the word “inconclusive” rather than saying the point is an inflection point. The test isn’t giving you a wrong answer; it’s giving you no answer.
Three Functions That Show Why
The classic way to see this is to compare three simple functions that all have f'(0) = 0 and f”(0) = 0, yet behave completely differently at the origin.
- f(x) = x³: The first derivative is 3x², which equals zero at x = 0. The second derivative is 6x, which also equals zero at x = 0. This point is neither a max nor a min. It’s an inflection point where the curve changes from concave down to concave up and passes right through without turning around.
- f(x) = x⁴: The first derivative is 4x³, zero at x = 0. The second derivative is 12x², also zero at x = 0. But this function does have a local minimum at the origin. The graph looks like a flattened bowl, sitting at y = 0 and curving upward on both sides.
- f(x) = −x⁴: Same logic, but flipped. Both first and second derivatives are zero at x = 0, and the point turns out to be a local maximum.
All three functions produce the same second derivative test result (zero), but one has a min, one has a max, and one has neither. That’s exactly why the test fails here: it can’t distinguish between these very different situations.
What to Do When the Test Fails
You have two reliable options: the first derivative test and the higher-order derivative test.
The First Derivative Test
Check the sign of f'(x) on both sides of the critical point. If the derivative switches from positive to negative, you have a local max. If it switches from negative to positive, you have a local min. If it doesn’t switch signs at all, the point is neither. This test never fails for continuous functions, which is why it’s the standard fallback.
For f(x) = x³ at x = 0, the derivative 3x² is positive on both sides of zero. No sign change means no extremum, confirming it’s just an inflection point. For f(x) = x⁴ at x = 0, the derivative 4x³ is negative to the left and positive to the right, confirming a local minimum.
The Higher-Order Derivative Test
If you’d rather stay in the “plug in and check” framework, you can keep taking derivatives until you find one that isn’t zero at the critical point. Suppose f'(c), f”(c), f”'(c), and so on are all zero, but the nth derivative is finally nonzero. Two rules determine what happens:
- If n is even and f⁽ⁿ⁾(c) > 0, the point is a local minimum.
- If n is even and f⁽ⁿ⁾(c) < 0, the point is a local maximum.
- If n is odd, the point is neither a max nor a min (it’s an inflection point).
For f(x) = x⁴, the third derivative is 24x, which is still zero at x = 0. The fourth derivative is 24, which is positive. Since 4 is even and the fourth derivative is positive, you’ve confirmed a local minimum. For f(x) = x³, the third derivative is 6, which is nonzero. Since 3 is odd, the point is an inflection point. The even/odd distinction makes intuitive sense: even-powered functions like x⁴ create symmetric bowl shapes, while odd-powered functions like x³ pass through the critical point without turning around.
Failure in Multivariable Calculus
For functions of two variables, the second derivative test uses a different criterion. At a critical point (x₀, y₀), you compute a value called the discriminant: AC − B², where A is the second partial derivative with respect to x twice, C is the second partial derivative with respect to y twice, and B is the mixed partial derivative. When AC − B² is positive or negative, the test gives a clear answer. When AC − B² equals exactly zero, the test fails for the same fundamental reason: the curvature information at that single point isn’t enough to determine the shape of the surface.
In the multivariable case, a failing test is harder to resolve. There’s no simple “first derivative test” equivalent in higher dimensions. You typically need to examine the function’s behavior along specific paths approaching the critical point, or use other analytical techniques to determine whether it’s a local extremum or a saddle point.
Recognizing the Pattern on Exams
In practice, the second derivative test fails most often on functions with repeated roots in their derivative, like x⁴, x⁶, or x³. Any time the critical point comes from a factor raised to a power of 2 or higher in f'(x), there’s a good chance f”(c) will also be zero. When you see that happening, skip straight to the first derivative test rather than spending time computing higher derivatives. It’s faster and always gives a definitive answer.