f(x) = tan^{-1} x - x is . . . . . . ,x in R | vedclass

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(C) To determine the nature of the function $f(x) = 	an^{-1} x - x$,we find its derivative with respect to $x$:$f'(x) = \frac{d}{dx}(	an^{-1} x - x)

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decreasing on $R$

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(C) To determine the nature of the function $f(x) = 	an^{-1} x - x$,we find its derivative with respect to $x$:$f'(x) = \frac{d}{dx}(	an^{-1} x - x) = \frac{1}{1+x^2} - 1$Simplifying the expression:$f'(x) = \frac{1 - (1+x^2)}{1+x^2} = \frac{-x^2}{1+x^2}$Since $x^2 \ge 0$ and $1+x^2 > 0$ for all $x \in R$,the derivative $f'(x) = \frac{-x^2}{1+x^2} \le 0$ for all $x \in R$.Since $f'(x) \le 0$ for all $x \in R$,the function $f(x)$ is a decreasing function on $R$.

$f(x) = \tan^{-1} x - x$ is . . . . . . ,$x \in R$.

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