Question

Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. ln(x^2+1)

Answer 1

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 $f^{l} (x)$ and equating zero

iii) solve and find 'x' values

iv) Find second derivative $f^{ll}(x) >0$ then find the minimum value at x=a

v) Find second derivative $f^{ll}(x) <0$ then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

step1:- find first derivative $f^{l} (x)$ and equating zero

  $f^{l}(x) = \frac{1}{x^2+1} \frac{d}{dx}(x^2+1)$

$f^{l}(x) = \frac{1}{x^2+1} (2x)$  ……………(1)

$f^{l}(x) = \frac{1}{x^2+1} (2x)=0$

the point is x=0

step2:-

Again differentiating with respective to 'x', we get

$f^{ll}(x)=\frac{x^2+1(2)-2x(2x)}{(x^2+1)^2}$

on simplification , we get

$f^{ll}(x) = \frac{-2x^2+2}{(x^2+1)^2}$

put x= 0 we get $f^{ll}(0) = \frac{2}{(1)^2}$   > 0

$f^{ll}(x) >0$ then find the minimum value at x=0

Final answer:-

The minimum value of the given function is f(0) = 0