Notes On Second Derivative Test - CBSE Class 12 Maths
Let f be a function defined on an interval I. Let a ∈ I be any element .Let f ''(a) exist. (i) If f '(a) = 0 and f ''(a) < 0, then a is the point of local maxima and f (a) is the local maximum value of f. (ii) If f '(a) = 0 and f ''(a) > 0, then a is the point of local minima and f (a) is the local minimum value of f. iii) If f '(a) = 0 and f ''(a) = 0, then the test fails. Given f '(a) = 0 f "(a) < 0  then f "(a) =  $\underset{h\to 0}{\text{lim}}\frac{\text{f'(a+h) - f'(a)}}{\text{h}}$  < 0    ⇒ f "(a) =  $\underset{h\to 0}{\text{lim}}\frac{\text{f'(a+h) - 0}}{\text{h}}$  < 0     ⇒ f "(a) =  $\underset{h\to 0}{\text{lim}}\frac{\text{f'(a+h)}}{\text{h}}$  < 0     f '(a + h)/h < 0 Case I: Suppose h < 0 ⇒ f '(a + h) > 0 Case II: Suppose h > 0 ⇒ f '(a + h) < 0 Hence, f has local maximum at point a. Local maximum value = f(a) If f '(a ) = 0 and f "(a) = 0, then the second derivative test fails. Working rule to find extreme values: Step1: Find all the points where f'(x) = 0 ⇒ x = x1,x2,x3,..... Step2: Find f "(x) Step3: Find f "(x1),f "(x2) and f "(x3).... Step4: If f "(x1) < 0, then the function has a local maximum value at x1. Local maximum value = f(x1) If f "(x1) > 0, then the function has a local minimum value at x1 . Local minimum value = f(x1).

#### Summary

Let f be a function defined on an interval I. Let a ∈ I be any element .Let f ''(a) exist. (i) If f '(a) = 0 and f ''(a) < 0, then a is the point of local maxima and f (a) is the local maximum value of f. (ii) If f '(a) = 0 and f ''(a) > 0, then a is the point of local minima and f (a) is the local minimum value of f. iii) If f '(a) = 0 and f ''(a) = 0, then the test fails. Given f '(a) = 0 f "(a) < 0  then f "(a) =  $\underset{h\to 0}{\text{lim}}\frac{\text{f'(a+h) - f'(a)}}{\text{h}}$  < 0    ⇒ f "(a) =  $\underset{h\to 0}{\text{lim}}\frac{\text{f'(a+h) - 0}}{\text{h}}$  < 0     ⇒ f "(a) =  $\underset{h\to 0}{\text{lim}}\frac{\text{f'(a+h)}}{\text{h}}$  < 0     f '(a + h)/h < 0 Case I: Suppose h < 0 ⇒ f '(a + h) > 0 Case II: Suppose h > 0 ⇒ f '(a + h) < 0 Hence, f has local maximum at point a. Local maximum value = f(a) If f '(a ) = 0 and f "(a) = 0, then the second derivative test fails. Working rule to find extreme values: Step1: Find all the points where f'(x) = 0 ⇒ x = x1,x2,x3,..... Step2: Find f "(x) Step3: Find f "(x1),f "(x2) and f "(x3).... Step4: If f "(x1) < 0, then the function has a local maximum value at x1. Local maximum value = f(x1) If f "(x1) > 0, then the function has a local minimum value at x1 . Local minimum value = f(x1).

Next