f(x) = cos x - 1 + frac{x^2}{2!}, x in R. The | vedclass

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(C) Given $f(x) = \cos x - 1 + \frac{x^2}{2!}$.Taking the derivative with respect to $x$, we get $f'(x) = -\sin x + x$.For $x > 0$, we know that $x >

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neither increasing nor decreasing

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(C) Given $f(x) = \cos x - 1 + \frac{x^2}{2!}$.Taking the derivative with respect to $x$, we get $f'(x) = -\sin x + x$.For $x > 0$, we know that $x > \sin x$, which implies $x - \sin x > 0$. Thus, $f'(x) > 0$ for $x > 0$.For $x Since the derivative $f'(x)$ changes sign at $x = 0$, the function $f(x)$ is neither increasing nor decreasing on the entire domain $R$.

$f(x) = \cos x - 1 + \frac{x^2}{2!}, x \in R$. Then $f(x)$ is

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