If f(x) = sin x - x cos x has a maximum at x | vedclass

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(D) Let $f(x) = \sin x - x \cos x$.First,find the derivative: $f'(x) = \cos x - (\cos x - x \sin x) = x \sin x$.For critical points,set $f'(x) = 0$,wh

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$n$ is an odd positive or even negative integer.

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(D) Let $f(x) = \sin x - x \cos x$.First,find the derivative: $f'(x) = \cos x - (\cos x - x \sin x) = x \sin x$.For critical points,set $f'(x) = 0$,which gives $x \sin x = 0$. Since $x = n\pi$,this is satisfied for any integer $n$.Next,find the second derivative: $f''(x) = \sin x + x \cos x$.Evaluate $f''(x)$ at $x = n\pi$: $f''(n\pi) = \sin(n\pi) + n\pi \cos(n\pi) = 0 + n\pi (-1)^n = n\pi (-1)^n$.For $f(x)$ to have a maximum at $x = n\pi$,we must have $f''(n\pi) Thus,$n\pi (-1)^n  0$,we require $n(-1)^n Case $1$: If $n$ is an odd positive integer (e.g.,$n=1, 3, 5$),then $n(-1)^n = n(-1) = -n Case $2$: If $n$ is an even negative integer (e.g.,$n=-2, -4$),then $n(-1)^n = n(1) = n Therefore,$n$ is an odd positive integer or an even negative integer.

If $f(x) = \sin x - x \cos x$ has a maximum at $x = n\pi$,then which of the following is true?

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