The function f(x) = cot^{-1} x + x is increas | vedclass

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(A) Given function is $f(x) = \cot^{-1} x + x$. To find the interval where the function is increasing,we calculate the derivative $f'(x)$. $f'(x) = \f

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$(-\infty, \infty)$

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(A) Given function is $f(x) = \cot^{-1} x + x$. To find the interval where the function is increasing,we calculate the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(\cot^{-1} x) + \frac{d}{dx}(x) = -\frac{1}{1+x^2} + 1$. Simplifying the expression: $f'(x) = \frac{-1 + (1+x^2)}{1+x^2} = \frac{x^2}{1+x^2}$. Since $x^2 \geq 0$ for all $x \in \mathbb{R}$ and $1+x^2 > 0$,it follows that $f'(x) \geq 0$ for all $x \in \mathbb{R}$. Since $f'(x) \geq 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is strictly increasing on the entire real line $(-\infty, \infty)$.

The function $f(x) = \cot^{-1} x + x$ is increasing in the interval.

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