Let f : R rightarrow R be defined as f(x) = x | vedclass

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(A) Given $f(x) = x^{3} + x - 5$.First,we find the derivative $f^{\prime}(x) = 3x^{2} + 1$.Since $f^{\prime}(x) > 0$ for all $x \in R$,the function $f

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$\frac{1}{49}$

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(A) Given $f(x) = x^{3} + x - 5$.First,we find the derivative $f^{\prime}(x) = 3x^{2} + 1$.Since $f^{\prime}(x) > 0$ for all $x \in R$,the function $f(x)$ is strictly increasing and therefore invertible.Given $f(g(x)) = x$,by the chain rule,we have $f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$,which implies $g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}$.To find $g^{\prime}(63)$,we need to find $x$ such that $f(x) = 63$.$x^{3} + x - 5 = 63 \Rightarrow x^{3} + x - 68 = 0$.By inspection,for $x = 4$,$4^{3} + 4 - 5 = 64 + 4 - 5 = 63$. Thus,$g(63) = 4$.Now,$g^{\prime}(63) = \frac{1}{f^{\prime}(g(63))} = \frac{1}{f^{\prime}(4)}$.Since $f^{\prime}(x) = 3x^{2} + 1$,we have $f^{\prime}(4) = 3(4)^{2} + 1 = 3(16) + 1 = 48 + 1 = 49$.Therefore,$g^{\prime}(63) = \frac{1}{49}$.

Let $f : R \rightarrow R$ be defined as $f(x) = x^{3} + x - 5$. If $g(x)$ is a function such that $f(g(x)) = x, \forall x \in R$,then $g^{\prime}(63)$ is equal to

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