Question

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

• A. $-1$
• B. $doesnotexist$
• C. 0
• D. 1

Answer 1

Answer: (A)

Given equation can be rewritten as
$f(x)=1-\frac{2}{x^2+1}$
Since, the maximum value of $\frac{2}{x^2+1}$ is $2$.
$\therefore$ Minimum value of $f(x)$ is $1-2=-1$