Question

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

• A. does not exist because f is unbounded
• 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 $-1$

$f\left(x\right)=\frac{x^2-1}{x^2+1}$
$f^{\prime }\left(x\right)=\frac{4x}{(x^2+1{)}^2}$
For maxima or minima,
$f^{\prime }\left(x\right)=0$
$\Rightarrow x=0$
$f^{\prime \prime }\left(x\right)>0$ at $x=0$
Hence f(x) has a minimum at $x=0$
$f\left(0\right)=-1$