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
