Second Derivative Test

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).