If $f(x)=\frac{x^2-1}{x^2+1}$, for every real number. Then what is the minimum value of f?
Answer & explanation
Correct answer: option 1
Let $f(x)=\frac{x^2-1}{x^2+1}$
$=\frac{x^2+1-2}{x^2+1}$
$=1-(\frac{2}{[x^2+1]})$ [Because $[x^2+1]>1$ also $(\frac{2}{[x^2+1]})≤2$]
So $1-[\frac{2}{[x^2+1]}]≥1-2$
$-1≤f(x)≤1$
Thus, f (x) has the minimum value equal to –1.