Second Derivative Test for Maxima and Minima

Here you will learn second derivative test for maxima and minima with examples.

Let’s begin –

Second Derivative Test for Maxima and Minima

If f(x) is continuous and differentiable at x = a where f'(a) = 0 and f”(a) also exists then for ascertaining maxima/minima at x = a, 2nd dervative can be used

(i) f”(a) > 0 $$\implies$$ x = a is a point of local minima

(ii) f”(a) < 0 $$\implies$$ x = a is a point of local maxima

(iii) f”(a) = 0 $$\implies$$ second derivative test fails.  To identify maxima/minima at this point either first derivative test or higher derivative test can be used.

Example : find all the points of local maxima and minima and the corresponding maximum and minimum values of the function f(x) = $$2x^3 – 21x^2 + 36x – 20$$.

Solution : We have,

f(x) = $$2x^3 – 21x^2 + 36x – 20$$

$$\implies$$ f'(x) = $$6x^2 – 42x +36$$

The critical points of f(x) are given by f'(x) = 0.

Now, f'(x) = 0 $$\implies$$ $$6x^2 – 42x +36$$

$$\implies$$ (x – 1)(x – 6) = 0 $$\implies$$ x = 1, 6.

Thus, x = 1 and x =6 are the possible points of local maxima and minima.

Now, we test the function at each of thes points.

We have, f”(x) = 12x – 42

At x = 1 : we have,

f”(1) = 12 – 42 = -30 < 0

So, x = 1 is a point of local maximum.

The local maximum value is f(1) = 2 – 21 + 36 – 20 = -3

At x = 6, We have,

f”(6) = 12(6) – 42 = 30 > 0

So, x = 6 is a point of local minimum.

The local minimum value is f(6) = -128