Maxima and Minima - Second Derivative Test

Take f(x) = x^3 - 3x^2 - 9x + 5. In the last article you found its critical points by solving f'(x) = 3(x-3)(x+1) = 0, getting x = -1 and x = 3. Then you built a sign chart for f', picked test values in each of three intervals, computed signs, and read off the classification: local max at x = -1, local min at x = 3.

Here is a much faster way. Compute the second derivative: f''(x) = 6x - 6. Plug in the critical points. f''(-1) = -12. Negative — bending down — so x = -1 is a local maximum. f''(3) = 12. Positive — bending up — so x = 3 is a local minimum. Done. No sign chart, no test values, no table — just two plug-ins.

This is the second derivative test. Instead of reading off how f' changes across a critical point, you read off how the curve is bending at the critical point. It is usually faster. It is always more satisfying. And when it works, you should use it.

The intuition, before the statement

Why should the second derivative know about maxima and minima?

The first derivative tells you about slope. At a critical point, the slope is zero — the tangent is horizontal. That alone does not tell you whether you're at a peak or a valley; a horizontal tangent happens at both.

The second derivative tells you about change in slope. If f''(c) < 0, the slope is decreasing at c. So just to the left of c, the slope was a bit higher than at c; just to the right, it will be a bit lower. But at c the slope is zero. So just to the left the slope is positive (the curve was going up), and just to the right the slope is negative (the curve is going down). That is the signature of a local maximum.

The exact same argument with inequalities flipped gives the local minimum case. If f''(c) > 0, the slope is increasing at c, so it went from negative (on the left) through zero (at c) to positive (on the right). The curve was going down, levelled off, and then went back up — a local minimum.

In picture form: f''(c) < 0 means the graph near c looks like the top of a frown; f''(c) > 0 means it looks like the bottom of a smile.

The whole test is just this picture made precise.

The test, stated and proved

Proof. Suppose f'(c) = 0 and f''(c) < 0. By definition,

since f'(c) = 0. The limit is negative. So for all sufficiently small |h|, the ratio \frac{f'(c+h)}{h} is also negative (it stays inside any neighbourhood of a negative number for h close enough to zero).

Read what that ratio being negative says. If h > 0, "ratio is negative" means f'(c + h) is negative — the slope is negative a tiny step to the right of c. If h < 0, the h in the denominator is negative, so for the ratio to be negative, the numerator f'(c + h) must be positive — the slope is positive a tiny step to the left of c.

Putting these together: f' is positive just to the left of c and negative just to the right. By the first derivative test, that is exactly the condition for a local maximum at c.

The f''(c) > 0 case is the same argument with inequalities reversed: f' is negative to the left of c and positive to the right, which by the first derivative test gives a local minimum. \blacksquare

The proof reveals something important. The second derivative test is not independent of the first derivative test — it is built out of the first derivative test. f''(c)'s sign controls whether f' is increasing or decreasing near c, which is exactly what you need to know.

So why bother with the second derivative test at all? Because plugging one number into f'' is much faster than building a sign chart for f' with test values. For polynomial problems, the speed-up is enormous. For problems with messy sign charts, it saves even more time.

Running the test

When the test is silent

The test is silent when f''(c) = 0. It does not say "there is no extremum" — it says "I cannot tell." And silence can be misleading, because f''(c) = 0 can happen in three very different situations:

• a local maximum (f(x) = -x^4 at x = 0 has f''(0) = 0, but it is still a max)
• a local minimum (f(x) = x^4 at x = 0 has f''(0) = 0, but it is still a min)
• neither (f(x) = x^3 at x = 0 has f''(0) = 0, and it is neither)

You can see all three in one picture.

When the second derivative test is silent, you have two choices:

1. Fall back to the first derivative test. Sign-chart f' near the critical point and read off the answer.
2. Use the higher-order derivative test — described next — which keeps differentiating until you find a derivative that is not zero at c.

For the three functions above:

• f(x) = x^4: f'(x) = 4x^3, which goes from negative (for x < 0) to positive (for x > 0). Local min. Matches the picture.
• f(x) = -x^4: f'(x) = -4x^3, which goes from positive to negative. Local max.
• f(x) = x^3: f'(x) = 3x^2, which is positive on both sides of 0. No sign change. Not an extremum.

The higher-order derivative test

Here is what "keep differentiating" means. Suppose f'(c) = 0 and f''(c) = 0 and f'''(c) = 0 and so on, up to some order n - 1, and the first nonzero derivative at c is f^{(n)}(c). Then Taylor's theorem says that near c,

The sign of the left-hand side — positive or negative — tells you whether the function sits above or below f(c) near the critical point, which is exactly what determines whether c is an extremum.

Look at the right-hand side. The factor \frac{f^{(n)}(c)}{n!} has a fixed sign (it is a single nonzero number). The factor (x - c)^n is the interesting one:

• if n is even, (x - c)^n \geq 0 always, with equality only at c. So the right side has constant sign on both sides of c. That means f(x) - f(c) has constant sign near c, which is the definition of an extremum.
• if n is odd, (x - c)^n is negative on one side and positive on the other. So f(x) - f(c) flips sign as you cross c, which means there is no extremum.

That gives a clean rule.

The ordinary second derivative test is the n = 2 case of this: n is even, so it's an extremum, and the sign of f''(c) picks max or min.

Apply it to the three examples where the second derivative test was silent:

• f(x) = x^4: f^{(4)}(x) = 24, so f^{(4)}(0) = 24. The first nonzero derivative is at order n = 4, which is even, and the derivative is positive. Local minimum.
• f(x) = -x^4: f^{(4)}(0) = -24. Order 4 is even, derivative is negative. Local maximum.
• f(x) = x^3: f^{(3)}(x) = 6, so f^{(3)}(0) = 6. Order n = 3 is odd. Not an extremum.

All three classifications match what the first derivative test said earlier. The higher-order test is just a cleaner way to reach the same answer — and it's particularly useful when you already have the Taylor series of a function.

Points of inflection

One important companion idea to the second derivative test: the point of inflection.

A point of inflection is a point on the graph where the concavity changes — the curve switches from bending down to bending up or vice versa. At such a point, f'' changes sign.

Two things to notice.

First, points of inflection are not the same as critical points. A critical point has f'(c) = 0 or undefined. A point of inflection has f'' changing sign at c. These are unrelated requirements. Many inflection points have nonzero f'. And many critical points have no inflection.

Second, f''(c) = 0 is necessary but not sufficient for an inflection. If f'' changes sign at c (and f is continuous), then f''(c) must equal zero (because a continuous function that changes sign must pass through zero). But f''(c) = 0 by itself does not mean a sign change — f(x) = x^4 has f''(0) = 0 but f'' is non-negative near 0, so no sign change, no inflection. Always follow up f''(c) = 0 with a sign check on f''.

Example: find the inflection points of f(x) = x^3 - 3x^2 + 1.

Compute f'': f'(x) = 3x^2 - 6x, f''(x) = 6x - 6. Set f''(x) = 0: x = 1. Check the sign of f'' on both sides: f''(0) = -6 < 0 (bending down), f''(2) = 6 > 0 (bending up). Sign change — so x = 1 is a point of inflection. The inflection point on the graph is (1, f(1)) = (1, -1).

You can see the curve switch from bending down (before x = 1) to bending up (after x = 1). That flip is the inflection point. For cubics, this always lives between the local max and local min, halfway along if the cubic is in standard form.

Common confusions

A few things that trip students up.

• "f''(c) = 0 means c is an inflection point." No — f''(c) = 0 is a necessary condition, not a sufficient one. You also need f'' to actually change sign at c. Always check both sides.

• "f''(c) = 0 means the test failed, so c is not an extremum." No — it means the test failed. The point might still be an extremum; you just need a different test. Fall back on the first derivative test or the higher-order test.

• Using the second derivative test at a non-critical point. The test only applies when f'(c) = 0. If f'(c) \neq 0, the concept of "local max" or "local min" at c doesn't make sense in this framework — the tangent isn't horizontal.

• Using the second derivative test on a non-differentiable critical point. If f'(c) does not exist, then neither does f''(c), and the second derivative test cannot be applied. Use the first derivative test.

• Confusing "concavity" with "extremum." Concavity is about how the curve is bending — a property that makes sense anywhere, not just at critical points. Extrema are about where the function hits local peaks or valleys. The two are related (at a local max, concavity is typically downward) but they are not the same.

Going deeper

Where the higher-order test comes from

The higher-order derivative test is really a consequence of Taylor's theorem. If f is n-times differentiable at c with f'(c) = \dots = f^{(n-1)}(c) = 0 and f^{(n)}(c) \neq 0, then Taylor's theorem with remainder gives

where the remainder R_n(x) satisfies \lim_{x \to c} \frac{R_n(x)}{(x - c)^n} = 0. So for x very close to c, the remainder is negligible compared to the main term, and

Everything follows from the sign analysis of (x - c)^n: even powers keep their sign, odd powers flip it.

This is worth internalising. The second derivative test is a first-order Taylor approximation. The higher-order derivative test extends it to functions that are too flat at c for the second derivative alone to catch.

When Taylor's theorem fails

There are pathological functions where every derivative at a point is zero but the function is not identically zero. The standard example is

Every derivative of f at 0 is zero. The Taylor series of f at 0 is identically zero, but f itself is not. Is 0 a local minimum of f? Yes — because f(x) > 0 for x \neq 0 and f(0) = 0. But the higher-order test can never detect this, because there is no "first nonzero derivative" to read off. You have to prove it from the definition directly.

You will not meet this in JEE. But it is a reminder that the machinery of derivatives, as powerful as it is, has a blind spot: functions that are "too flat" at a point are invisible to every derivative-based test. These are called non-analytic functions, and they are exactly the gap between the real-valued theory of calculus and the complex-valued theory, where such gaps cannot exist.

The genuinely subtle case with inflection and the first derivative

One last thing. A point where f''(c) = 0 and f'' does change sign is an inflection point. But what happens to f' at such a point? f' can take any value — it might even be zero.

If f''(c) = 0, f'' changes sign at c, and f'(c) = 0, then c is simultaneously a critical point and an inflection point. That is the x^3 case: f'(0) = 0 (critical), f''(0) = 0, and f'' changes sign at 0 (inflection). The net result is that the tangent at c is horizontal but the curve still passes through — a so-called horizontal point of inflection. These are the critical points that the higher-order test classifies as "not an extremum."

Where this leads next

The first and second derivative tests are the core toolkit for finding extrema. The next articles combine them with the rest of the geometric information about a curve.

• Concavity and Points of Inflection — a deeper look at concavity on its own, without the pretext of finding extrema.
• Curve Sketching — combining extrema, concavity, intercepts, and asymptotes to draw a full graph by hand.
• Optimization Problems — "largest / smallest" word problems, where the hardest part is setting up the function to optimize.
• Maxima and Minima - Special Cases — non-differentiable functions, endpoint extrema, and the global-vs-local distinction.
• Maxima and Minima - First Derivative Test — the companion test, still needed whenever the second derivative test is silent.