Answer to Question #204599 in Calculus for ali

Question

Determine the location of all critical points and determine their nature for the function. f(x)=ln(x^2 + 1) - x

Answer 1

Here $f(x)=ln(x^2+1)-x$

We need to find location of all critical points & their nature for the function. For that we need to put $f'(x)=0$

Here $f'(x)=\frac {d} {dx} (ln(x^2+1)-x)$

$=\frac {2x} {x^2+1}$ $-1$ $\because \frac {d} {dx} (ln (x))=\frac 1 x$ & $\frac {d} {dx} (x^n)=nx^{n-1}$

To find critical points we will put $f'(x)=0$

So $\frac {2x} {x^2+1} -1=0$

$\implies \frac {2x} {x^2+1} =1$

$\implies 2x=x^2+1$

$\implies x^2-2x+1=0$

$\implies (x-1)^2=0$

$\implies x=1$

So at $x=1$ we get critical point & when $x=1$ , $f(1)=ln(1+1)-1=ln2-1$

So critical point is $(1,ln2 -1)$

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

$\because \frac {d} {dx} (\frac u v)=\frac {(v.u'-u.v')} {v^2}$ & $\frac {d} {dx} (constant)=0$

Now $f''(1)=\frac {(2-2)} {(2)^2} =\frac 0 4 =0$

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