Let f(x)=x^5+2x^3+3x+1,x in R,and g(x) be a f | vedclass

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(D) Given $f(x)=x^5+2x^3+3x+1$.First,we find the derivative $f^{\prime}(x) = 5x^4+6x^2+3$.We are given $g(f(x))=x$. Differentiating both sides with re

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$14$

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(D) Given $f(x)=x^5+2x^3+3x+1$.First,we find the derivative $f^{\prime}(x) = 5x^4+6x^2+3$.We are given $g(f(x))=x$. Differentiating both sides with respect to $x$ using the chain rule,we get $g^{\prime}(f(x)) \cdot f^{\prime}(x) = 1$.To find $g(7)$ and $g^{\prime}(7)$,we set $f(x)=7$:$x^5+2x^3+3x+1=7 \Rightarrow x^5+2x^3+3x-6=0$.By inspection,$x=1$ is a solution since $1+2+3-6=0$.Thus,$f(1)=7$,which implies $g(7)=1$.Now,substitute $x=1$ into the derivative equation $g^{\prime}(f(1)) \cdot f^{\prime}(1) = 1$:$g^{\prime}(7) \cdot f^{\prime}(1) = 1$.We calculate $f^{\prime}(1) = 5(1)^4+6(1)^2+3 = 5+6+3 = 14$.Therefore,$g^{\prime}(7) = \frac{1}{f^{\prime}(1)} = \frac{1}{14}$.Finally,$\frac{g(7)}{g^{\prime}(7)} = \frac{1}{1/14} = 14$.

Let $f(x)=x^5+2x^3+3x+1$,$x \in R$,and $g(x)$ be a function such that $g(f(x))=x$ for all $x \in R$. Then $\frac{g(7)}{g^{\prime}(7)}$ is equal to:

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