Question

If $f\left(x\right)=\frac{x^2-1}{x^2+1}$, for every real number, then minimum value of $f\left(x\right)$

• A. Does not exist
• B. Is not attained even through $f$ is bounded
• C. Is equal to $1$
• D. Is equal to $-1$

Answer 1

The correct option is D Is equal to $-1$

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

$f^{\prime }\left(x\right)==\frac{(x^2+1)2x-({x1}^2-)\left(2x\right)}{{(x^2+1)}^2}=\frac{2x^3+2x-2x^3+2x}{{(x^2+1)}^2}=\frac{4x}{{(x^2+1)}^2}$

$f^{\prime }\left(x\right)=0$

$\Rightarrow$ $x=0$

If $x<0,f^{\prime }\left(x\right)<0$

If $x>0$; $f^{\prime }\left(x\right)>0$

$\therefore$ $x$ is the point of minima

Minimum value of $f\left(x\right)=f\left(0\right)=-1$

$\therefore$ Minimum value of $f\left(x\right)=1$