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

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(D) Let $f(x) = \frac{x^2 - 1}{x^2 + 1}$.We can rewrite the function as:$f(x) = \frac{x^2 + 1 - 2}{x^2 + 1} = 1 - \frac{2}{x^2 + 1}$.Since $x^2 \ge 0$

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Is equal to $-1$

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(D) Let $f(x) = \frac{x^2 - 1}{x^2 + 1}$.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 numbers $x$,we have $x^2 + 1 \ge 1$.This implies that $0 Multiplying by $2$,we get $0 Now,multiplying by $-1$ reverses the inequality:$-2 \le -\frac{2}{x^2 + 1} Adding $1$ to all parts:$1 - 2 \le 1 - \frac{2}{x^2 + 1} $-1 \le f(x) Thus,the minimum value of $f(x)$ is $-1$,which is attained when $x = 0$.

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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