Question

Determine the location of all critical points and determine the nature for the function f(x)=In(x^2+1)_x.​

Answer 1

$\\ \bigstar \: \large \underline \purple{ \rm{Given :- }}$

• $\sf \: f(x) = In( {x}^{2} + 1) - x$

$\\ \bigstar \: \: \large \underline \purple{ \rm{To \: find :- }}$

• The location of all critical points and their nature of the function.

$\\ \bigstar \: \: \large \underline \purple{ \rm{Solution :- }}$

In this question, We have to find the location of all critical points and their nature of the function.

So, for this we have to put f'(x) = 0

Here,

$\\ \sf \: \: f'(x) = \frac{d}{dx} [In( {x}^{2} + 1) - x]$

As we know,

$\\ \: \: \: \: \: \: \: \: \sf \: \: \frac{d}{dx} [In( {x}) ] = \frac{1}{x} \: and \: \frac{d}{dx} ( {x}^{n}) = n {x}^{n - 1}$

So,

$\\ \: \: \: \: \: \: \: \: \: \implies \: \sf \frac{2x}{ {x}^{2} + 1} - 1$

Now ,to find critical points let's put f'(x)=0

Therefore,

$\\ \: \: \: \: \: \: \: \: \: \qquad \: \sf \frac{2x}{ {x}^{2} + 1} - 1 = 0$

$\\ \: \: \: \: \: \: \: \: \: \longrightarrow \: \ \sf \frac{2x}{ {x}^{2} + 1} = 1$

$\\ \: \: \: \: \: \: \: \: \: \longrightarrow \: \ \sf {2x} = 1( {x}^{2} + 1)$

$\\ \: \: \: \: \: \: \: \: \: \longrightarrow \: \ \sf {2x} = ( {x}^{2} + 1)$

$\\ \: \: \: \: \: \: \: \: \: \longrightarrow \: \ \sf {x}^{2} - 2x + 1 = 0$

$\\ \: \: \: \: \: \: \: \: \: \longrightarrow \: \ \sf {(x - 1)}^{2} = 0$

$\\ \: \: \: \: \: \: \: \: \: \: \ \sf \: \: \dots \: \: because \: \: {a}^{2} - 2ab + {b}^{2} = {(a - b)}^{2}$

Therefore ,

$\\ \: \: \: \: \: \: \: \: \: \longrightarrow \: \ \sf {x} = 1$

So, at x = 1 we get critical point and when x = 1 ,

$\\ \implies\sf \: f(1) = In( {1}^{2} + 1) - 1$

$\\ \implies\sf \: f(1) = In( {1}+ 1) - 1$

$\\ \implies\sf \: f(1) = In2- 1$

Therefore, critical point is ( 1, In2-1)

Now,

$\\ \sf \: \: f''(x) = \frac{d}{dx} \bigg( \frac{2x}{ {x}^{2} + 1 } \bigg)$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: = \frac{2 - {2x}^{2} }{( {x}^{2} + 1)^{2} }$

$\\ \: \: \: \: \: \: \: \: \: \: \ \sf \: \: \dots \: \: because \: \: \: \: \frac{d}{dx} \bigg( \frac{u}{v} \bigg) = \frac{(v.u' -u.v')}{ {v}^{2} } \: and \: \frac{d}{dx} = 0$

Now,

$\\ \sf \: \: f''(1) = \frac{(2 - 2)}{ {(2)}^{2} }$

$\\ \sf \: \: \: \: \: \: \: \: \: = \frac{0}{4 } = 0$

As f''(1) = 0 So, we will get inflection point x = 1. That is there is no local Maxima and local minima.