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

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(D) Given $f(x) = \frac{x^2 - 1}{x^2 + 1}$.We can rewrite the function as $f(x) = \frac{x^2 + 1 - 2}{x^2 + 1} = 1 - \frac{2}{x^2 + 1}$.Since $x^2 \ge

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Is equal to $-1$

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(D) Given $f(x) = \frac{x^2 - 1}{x^2 + 1}$.We can rewrite the function as $f(x) = \frac{x^2 + 1 - 2}{x^2 + 1} = 1 - \frac{2}{x^2 + 1}$.Since $x^2 \ge 0$ for all real $x$,we have $x^2 + 1 \ge 1$,which implies $0 Subtracting this inequality from $1$,we get $1 - 2 \le 1 - \frac{2}{x^2 + 1} Thus,$-1 \le f(x) The minimum value is attained when $x^2 = 0$,i.e.,$x = 0$.$f(0) = \frac{0^2 - 1}{0^2 + 1} = -1$.Therefore,the minimum value of $f$ is $-1$.

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

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