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(C) Given the function $f(x) = x + \sin x$. First,we find the derivative: $f'(x) = 1 + \cos x$. Setting $f'(x) = 0$,we get $1 + \cos x = 0$,which impl

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Neither maximum nor minimum

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(C) Given the function $f(x) = x + \sin x$. First,we find the derivative: $f'(x) = 1 + \cos x$. Setting $f'(x) = 0$,we get $1 + \cos x = 0$,which implies $\cos x = -1$. This occurs at $x = (2n+1)\pi$ for any integer $n$. Now,we check the second derivative: $f''(x) = -\sin x$. At $x = \pi$,$f''(\pi) = -\sin(\pi) = 0$. Since the second derivative is zero,we check the third derivative: $f'''(x) = -\cos x$. At $x = \pi$,$f'''(\pi) = -\cos(\pi) = -(-1) = 1 
eq 0$. Since the first non-zero derivative at the critical point is of an odd order (the third derivative),the function $f(x)$ has a point of inflection at $x = \pi$ and does not have a local maximum or local minimum.

The function $f(x) = x + \sin x$ has

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