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(C) We are given $g(x) = x f^{\prime}(x)$.To find $g^{\prime}(0)$, we use the definition of the derivative:$g^{\prime}(0) = \lim_{h 	o 0} \frac{g(0+h

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$g^{\prime}(0)=1$

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(C) We are given $g(x) = x f^{\prime}(x)$.To find $g^{\prime}(0)$, we use the definition of the derivative:$g^{\prime}(0) = \lim_{h 	o 0} \frac{g(0+h) - g(0)}{h}$.Since $g(0) = 0 \cdot f^{\prime}(0) = 0$, we have:$g^{\prime}(0) = \lim_{h 	o 0} \frac{h f^{\prime}(h) - 0}{h} = \lim_{h 	o 0} f^{\prime}(h)$.Since $f^{\prime}(x)$ is given to be continuous at $x=0$, we have $\lim_{h 	o 0} f^{\prime}(h) = f^{\prime}(0)$.Given $f^{\prime}(0) = 1$, it follows that $g^{\prime}(0) = 1$.

Suppose that $f(x)$ is a differentiable function such that $f^{\prime}(x)$ is continuous, $f^{\prime}(0)=1$ and $f^{\prime \prime}(0)$ does not exist. Let $g(x)=x f^{\prime}(x)$. Then,

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