The points of extrema of f(x) = int_0^x frac{ | vedclass

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(D) Given the function $f(x) = \int_0^x \frac{\sin t}{t} dt$. By the Fundamental Theorem of Calculus,the derivative is $f'(x) = \frac{\sin x}{x}$. For

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$n\pi; n = 1, 2, \dots$

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(D) Given the function $f(x) = \int_0^x \frac{\sin t}{t} dt$. By the Fundamental Theorem of Calculus,the derivative is $f'(x) = \frac{\sin x}{x}$. For extrema,we set $f'(x) = 0$. $\frac{\sin x}{x} = 0$ implies $\sin x = 0$ for $x > 0$. This occurs at $x = n\pi$ for $n = 1, 2, 3, \dots$. Thus,the points of extrema are $x = n\pi$ where $n \in \mathbb{N}$.

The points of extrema of $f(x) = \int_0^x \frac{\sin t}{t} dt$ in the domain $x > 0$ are

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