Answer:
The minimum value of the given function is f(0) = 0
Step-by-step explanation:
Explanation:-
Extreme value :- f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.
i) the necessary and sufficient condition for f(x) to have a maximum or minimum at given point.
ii) find first derivative
and equating zero
iii) solve and find 'x' values
iv) Find second derivative
then find the minimum value at x=a
v) Find second derivative
then find the maximum value at x=a
Problem:-
Given function is f(x) = log ( x^2 +1)
<u>step1:</u>- find first derivative
and equating zero

……………(1)

the point is x=0
<u>step2:-</u>
Again differentiating with respective to 'x', we get

on simplification , we get

put x= 0 we get
> 0
then find the minimum value at x=0
<u>Final answer</u>:-
The minimum value of the given function is f(0) = 0