The function f(x) = x + cos x is | vedclass

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(A) Given the function $f(x) = x + \cos x$. To determine the nature of the function,we find its derivative with respect to $x$: $f'(x) = \frac{d}{dx}(

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Always increasing

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(A) Given the function $f(x) = x + \cos x$. To determine the nature of the function,we find its derivative with respect to $x$: $f'(x) = \frac{d}{dx}(x + \cos x) = 1 - \sin x$. We know that for any real number $x$,the range of $\sin x$ is $[-1, 1]$. Therefore,$1 - \sin x \ge 0$ for all $x \in \mathbb{R}$. Specifically,$f'(x) = 0$ only at points where $\sin x = 1$ (i.e.,$x = 2n\pi + \frac{\pi}{2}$),and $f'(x) > 0$ elsewhere. Since $f'(x) \ge 0$ for all $x$ and does not vanish on any interval,the function $f(x)$ is strictly increasing on $\mathbb{R}$.

The function $f(x) = x + \cos x$ is

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