For the function f(x) = int_{0}^{x} frac{sin | vedclass

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(C) Given $f(x) = \int_{0}^{x} \frac{\sin t}{t} dt$.By the Fundamental Theorem of Calculus,$f'(x) = \frac{\sin x}{x}$ for $x > 0$.To find critical poi

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It has a maximum at $x = n\pi$,where $n$ is odd.

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(C) Given $f(x) = \int_{0}^{x} \frac{\sin t}{t} dt$.By the Fundamental Theorem of Calculus,$f'(x) = \frac{\sin x}{x}$ for $x > 0$.To find critical points,set $f'(x) = 0$,which implies $\sin x = 0$,so $x = n\pi$ for $n = 1, 2, 3, \dots$.The second derivative is $f''(x) = \frac{x \cos x - \sin x}{x^2}$.At $x = n\pi$,$f''(n\pi) = \frac{n\pi \cos(n\pi) - \sin(n\pi)}{(n\pi)^2} = \frac{n\pi (-1)^n - 0}{(n\pi)^2} = \frac{(-1)^n}{n\pi}$.If $n$ is odd,$f''(n\pi) = \frac{-1}{n\pi} If $n$ is even,$f''(n\pi) = \frac{1}{n\pi} > 0$,which indicates a local minimum.Thus,the function has a maximum at $x = n\pi$ where $n$ is odd.

For the function $f(x) = \int_{0}^{x} \frac{\sin t}{t} dt$,where $x > 0$,which of the following is true?

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