Let g(x) = int_{x}^{2x} frac{f(t)}{t} dt wher | vedclass

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(C) Given $g(x) = \int_{x}^{2x} \frac{f(t)}{t} dt$.Using the Leibniz integral rule, we differentiate $g(x)$ with respect to $x$:$g'(x) = \frac{f(2x)}{

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$g(x)$ is a constant function

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(C) Given $g(x) = \int_{x}^{2x} \frac{f(t)}{t} dt$.Using the Leibniz integral rule, we differentiate $g(x)$ with respect to $x$:$g'(x) = \frac{f(2x)}{2x} \cdot \frac{d}{dx}(2x) - \frac{f(x)}{x} \cdot \frac{d}{dx}(x)$$g'(x) = \frac{f(2x)}{2x} \cdot 2 - \frac{f(x)}{x} \cdot 1$$g'(x) = \frac{f(2x)}{x} - \frac{f(x)}{x}$Since it is given that $f(2x) = f(x)$, we substitute this into the expression:$g'(x) = \frac{f(x) - f(x)}{x} = 0$Since $g'(x) = 0$ for all $x > 0$, the function $g(x)$ is a constant function.

Let $g(x) = \int_{x}^{2x} \frac{f(t)}{t} dt$ where $x > 0$ and $f$ is a continuous function such that $f(2x) = f(x)$. Then:

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