{"id":5672,"date":"2021-09-27T19:46:23","date_gmt":"2021-09-27T14:16:23","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5672"},"modified":"2021-11-13T17:12:10","modified_gmt":"2021-11-13T11:42:10","slug":"second-derivative-test-for-maxima-and-minima","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/second-derivative-test-for-maxima-and-minima\/","title":{"rendered":"Second Derivative Test for Maxima and Minima"},"content":{"rendered":"
Here you will learn second derivative test for maxima and minima with examples.<\/p>\n
Let’s begin –<\/p>\n
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<\/p>\n
\n(i) f”(a) > 0 \\(\\implies\\) x = a is a point of local minima<\/p>\n
(ii) f”(a) < 0 \\(\\implies\\) x = a is a point of local maxima<\/p>\n
(iii) f”(a) = 0 \\(\\implies\\) second derivative test fails.\u00a0 To identify maxima\/minima at this point either first derivative test or higher derivative test can be used.<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : 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\\).<\/p>\n
Solution<\/strong><\/span> : We have,\u00a0<\/p>\n
f(x) = \\(2x^3 – 21x^2 + 36x – 20\\)<\/p>\n
\\(\\implies\\) f'(x) = \\(6x^2 – 42x +36\\)<\/p>\n
The critical points of f(x) are given by f'(x) = 0.<\/p>\n
Now, f'(x) = 0 \\(\\implies\\) \\(6x^2 – 42x +36\\)\u00a0<\/p>\n
\\(\\implies\\) (x – 1)(x – 6) = 0 \\(\\implies\\) x = 1, 6.<\/p>\n
Thus, x = 1 and x =6 are the possible points of local maxima and minima.<\/p>\n
Now, we test the function at each of thes points.<\/p>\n
We have, f”(x) = 12x – 42<\/p>\n
At x = 1 : we have,<\/p>\n
f”(1) = 12 – 42 = -30 < 0<\/p>\n
So, x = 1 is a point of local maximum.<\/p>\n
The local maximum value is f(1) = 2 – 21 + 36 – 20 = -3<\/p>\n
At x = 6, We have,<\/p>\n
f”(6) = 12(6) – 42 = 30 > 0<\/p>\n
So, x = 6 is a point of local minimum.<\/p>\n
The local minimum value is f(6) = -128<\/p>\n\n\n