Let f: R rightarrow R be a differentiable fun | vedclass

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(A) For $g(x)$ to be continuous at $x=3$,we must have $K = \lim_{x 	o 3} g(x) = \lim_{x 	o 3} \int_3^{f(x)} \frac{3t^2}{x-3} dt$. Since the integral

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(A) For $g(x)$ to be continuous at $x=3$,we must have $K = \lim_{x 	o 3} g(x) = \lim_{x 	o 3} \int_3^{f(x)} \frac{3t^2}{x-3} dt$. Since the integral is of the form $\frac{0}{0}$ as $x 	o 3$ (because $f(3)=3$),we apply $L$'$H$ôpital's Rule and Leibniz's Rule for differentiation under the integral sign. Let $I(x) = \int_3^{f(x)} 3t^2 dt$. Then $g(x) = \frac{I(x)}{x-3}$. By $L$'$H$ôpital's Rule,$K = \lim_{x 	o 3} \frac{\frac{d}{dx} \int_3^{f(x)} 3t^2 dt}{\frac{d}{dx} (x-3)}$. Using Leibniz's Rule,$\frac{d}{dx} \int_3^{f(x)} 3t^2 dt = 3(f(x))^2 \cdot f^{\prime}(x)$. Thus,$K = \lim_{x 	o 3} \frac{3(f(x))^2 \cdot f^{\prime}(x)}{1} = 3(f(3))^2 \cdot f^{\prime}(3)$. Substituting the given values $f(3)=3$ and $f^{\prime}(3)=\frac{1}{27}$: $K = 3 \cdot (3)^2 \cdot \frac{1}{27} = 3 \cdot 9 \cdot \frac{1}{27} = 27 \cdot \frac{1}{27} = 1$.

Let $f: R \rightarrow R$ be a differentiable function such that $f(3)=3$ and $f^{\prime}(3)=\frac{1}{27}$. If $g(x)=\begin{cases} \int_3^{f(x)} \frac{3t^2}{x-3} dt & \text{if } x \neq 3 \\ K & \text{if } x=3 \end{cases}$ is continuous at $x=3$,then $K=$

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