Question

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

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

for finding minimum value or maximum value of $f$,
we have to calculate $f^{\prime }\left(x\right)=0$
$f^{\prime }\left(x\right)=\frac{4x}{(x^2+1)}=0=>x=0$
$f^{\prime \prime }\left(x\right)$ at $x=0$ is positive thus at $x=0$ will get minimum value.
for $x=0,f$ will be minimum $=-1$