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  4. If f(x) = (x2-1/x2+1) for every real number x, then the minimum value of f is

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

AMUAMU 2011Relations and Functions

Solution:

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$