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
![f^{l}(x) = \frac{1}{x^2+1} \frac{d}{dx}(x^2+1)](https://tex.z-dn.net/?f=f%5E%7Bl%7D%28x%29%20%3D%20%5Cfrac%7B1%7D%7Bx%5E2%2B1%7D%20%5Cfrac%7Bd%7D%7Bdx%7D%28x%5E2%2B1%29)
……………(1)
![f^{l}(x) = \frac{1}{x^2+1} (2x)=0](https://tex.z-dn.net/?f=f%5E%7Bl%7D%28x%29%20%3D%20%5Cfrac%7B1%7D%7Bx%5E2%2B1%7D%20%282x%29%3D0)
the point is x=0
<u>step2:-</u>
Again differentiating with respective to 'x', we get
![f^{ll}(x)=\frac{x^2+1(2)-2x(2x)}{(x^2+1)^2}](https://tex.z-dn.net/?f=f%5E%7Bll%7D%28x%29%3D%5Cfrac%7Bx%5E2%2B1%282%29-2x%282x%29%7D%7B%28x%5E2%2B1%29%5E2%7D)
on simplification , we get
![f^{ll}(x) = \frac{-2x^2+2}{(x^2+1)^2}](https://tex.z-dn.net/?f=f%5E%7Bll%7D%28x%29%20%3D%20%5Cfrac%7B-2x%5E2%2B2%7D%7B%28x%5E2%2B1%29%5E2%7D)
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