Question

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

• A. does not exist
• B. is not attained even though $f$ is bounded
• C. is equal to $1$
• D. is equal to $-1$

Answer 1

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

We have $f\left(x\right)=1-\frac{2}{x^2+1}$
$f\left(x\right)$ will be minimum if $2/(x^2+1)$ is maximum i.e. if $x^2+1$ is least i.e. when $x=0$.
Thus, minimum value of $f\left(x\right)$ is $f\left(0\right)=-1$.