Second Derivative Test for Maxima and Minima

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

Letโ€™s begin โ€“

Second Derivative Test for Maxima and Minima

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

(i) fโ€(a) > 0 \(\implies\) x = a is a point of local minima

(ii) fโ€(a) < 0 \(\implies\) x = a is a point of local maxima

(iii) fโ€(a) = 0 \(\implies\) second derivative test fails.ย  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) = \(2x^3 โ€“ 21x^2 + 36x โ€“ 20\).

Solution : We have,ย 

f(x) = \(2x^3 โ€“ 21x^2 + 36x โ€“ 20\)

\(\implies\) f'(x) = \(6x^2 โ€“ 42x +36\)

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

Now, f'(x) = 0 \(\implies\) \(6x^2 โ€“ 42x +36\)ย 

\(\implies\) (x โ€“ 1)(x โ€“ 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, fโ€(x) = 12x โ€“ 42

At x = 1 : we have,

fโ€(1) = 12 โ€“ 42 = -30 < 0

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

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

At x = 6, We have,

fโ€(6) = 12(6) โ€“ 42 = 30 > 0

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

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

Leave a Comment

Your email address will not be published. Required fields are marked *

Ezoicreport this ad