Let f(x) = intlimits_0^x frac{sin t}{t} dt fo | vedclass

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(D) Given $f(x) = \int_0^x \frac{\sin t}{t} dt$. By the Fundamental Theorem of Calculus,$f'(x) = \frac{\sin x}{x}$. For critical points,set $f'(x) = 0

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Both $(A)$ and $(B)$

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(D) Given $f(x) = \int_0^x \frac{\sin t}{t} dt$. By the Fundamental Theorem of Calculus,$f'(x) = \frac{\sin x}{x}$. For critical points,set $f'(x) = 0$,which implies $\sin x = 0$ (since $x > 0$),so $x = n\pi$ for $n = 1, 2, 3, \dots$. Now,$f''(x) = \frac{x \cos x - \sin x}{x^2}$. At $x = n\pi$,$f''(n\pi) = \frac{n\pi \cos(n\pi) - \sin(n\pi)}{(n\pi)^2} = \frac{n\pi (-1)^n - 0}{n^2 \pi^2} = \frac{(-1)^n}{n\pi}$. If $n$ is odd $(n = 1, 3, 5, \dots)$,$f''(n\pi) = \frac{-1}{n\pi} If $n$ is even $(n = 2, 4, 6, \dots)$,$f''(n\pi) = \frac{1}{n\pi} > 0$,so $f(x)$ has a local minimum. Thus,both $(A)$ and $(B)$ are correct.

Let $f(x) = \int\limits_0^x \frac{\sin t}{t} dt$ for $x > 0$. Then $f(x)$ has:

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