If f(x) = frac{{{x^2} - 1}}{{{x^2} + 1}}, for | vedclass

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Question

(d) 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}

Vedclass · Accepted Answer

Is equal to $-1$

Vedclass · Answer

(d) Let $f(x) = \frac{{{x^2} - 1}}{{{x^2} + 1}} $

$= \frac{{{x^2} + 1 - 2}}{{{x^2} + 1}} = 1 - \frac{2}{{{x^2} + 1}}$

$\because {{x^2} + 1} > 1;   \,\, 	herefore  \frac{2}{{{x^2} + 1}} \le 2$

$	herefore  1 - \frac{2}{{{x^2} + 1}} \ge 1 - 2$; 

$	herefore - 1 \le f(x) < 1$

Thus $f(x)$ has the minimum value equal to $-1$.

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

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