Second derivative test

Let f be twice differentiable at c with f′(c) = 0. If f″(c) < 0, c is a local maximum; if f″(c) > 0, c is a local minimum.

Picture: f″ < 0 means the slope is falling, so the curve bends down like a cap; f″ > 0 means it bends up like a cup.

If f″(c) = 0 the test fails and says nothing; go back to the first derivative test.

Worked example: f(x) = 3x⁴ + 4x³ − 12x² + 12. f′ = 12x(x − 1)(x + 2) and f″ = 12(3x² + 2x − 2). f″(0) = −24 gives a local maximum f(0) = 12; f″(1) = 36 and f″(−2) = 72 give local minima f(1) = 7 and f(−2) = −20.

Worked example: f(x) = 2x³ − 6x² + 6x + 5 has f′(1) = 0 and f″(1) = 0. The second test fails, and the first test shows no extremum at 1.

Worked example: two positive numbers add to 15. The sum of squares S = x² + (15 − x)² has S′ = 4x − 30 = 0 at x = 15/2 and S″ = 4 > 0, so 15/2 and 15/2 give the least sum of squares.