{"id":5670,"date":"2021-09-27T19:45:17","date_gmt":"2021-09-27T14:15:17","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5670"},"modified":"2021-11-13T17:27:37","modified_gmt":"2021-11-13T11:57:37","slug":"first-derivative-test-for-maxima-and-minima","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/first-derivative-test-for-maxima-and-minima\/","title":{"rendered":"First Derivative Test for Maxima and Minima"},"content":{"rendered":"
Here you will learn first derivative test for maxima and minima with examples.<\/p>\n
Let’s begin –<\/p>\n
If f'(x) = 0 at a point (say x = a) and <\/p>\n
\n(i) If f'(x) changes sign from positive to negative in the neighbourhood of x = a then x = a is said to be a point local maxima.<\/strong><\/p>\n
(ii) If f'(x) changes sign from negative to positive in the neighbourhood of x = a then x = a is said to be a point local minima.<\/strong><\/p>\n<\/blockquote>\n
Note<\/strong> – If f'(x) does not change sign i.e. has the same sign in a certain complete neighbourhood of a, then f(x) is either increasing or decreasing throughout this neighbourhood implying that x = a is not a point of extremum of f.<\/p>\n
Example<\/strong><\/span> : Let f(x) = x + \\(1\\over x\\) ; x \\(\\ne\\) 0. Discuss the local maximum and local minimum values of f(x)<\/p>\n
Solution<\/span><\/strong> : Here f'(x) = 1 – \\(1\\over x^2\\) = \\(x^2 – 1\\over x^2\\)<\/p>\n
= \\((x – 1)(x + 1)\\over x^2\\)<\/p>\n
Using local number line rule, <\/p>\n
f(x) will have local maximum at x = – 1 and local minimum at x =1 <\/p>\n
\\(\\therefore\\) local maximum value of f(x) = -2 at x = -1<\/p>\n
and local minimum value of f(x) = 2 at x = 1<\/p>\n
Example<\/strong><\/span> : Let f(x) = \\(x^3 – 3x\\). Discuss the local maximum and local minimum values of f(x)<\/p>\n
Solution<\/span><\/strong> : Here f'(x) = \\(3x^2 -3\\) = 3(x + 1)(x – 1)<\/p>\n
Using local number line rule, <\/p>\n
f(x) will have local maximum at x = – 1 and local minimum at x =1 <\/p>\n
\\(\\therefore\\) local maximum value of f(x) = 2 at x = -1<\/p>\n
and local minimum value of f(x) = -2 at x = 1<\/p>\n\n\n