Let f(x) = e^x - e^{-x} + cos x,then f(x) is | vedclass

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(A) Given $f(x) = e^x - e^{-x} + \cos x$. First,we find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(e^x - e^{-x} + \cos x) = e^x + e^{-x} - \sin x$.

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always increasing

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(A) Given $f(x) = e^x - e^{-x} + \cos x$. First,we find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(e^x - e^{-x} + \cos x) = e^x + e^{-x} - \sin x$. We know that $e^x + e^{-x} = 2 \cosh x$. Since $e^x + e^{-x} \ge 2$ for all $x \in \mathbb{R}$ (by $AM$-$GM$ inequality) and $\sin x \le 1$ for all $x \in \mathbb{R}$,we have: $f'(x) = (e^x + e^{-x}) - \sin x \ge 2 - 1 = 1$. Since $f'(x) > 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is always increasing.

Let $f(x) = e^x - e^{-x} + \cos x$,then $f(x)$ is

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