Let f(x)=x^5+2x^3+3x+1,x R, and g(x) be a function such that g(f(x))=x for all x R.

Question

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

Accepted Answer

$\begin{gathered} f(x)={x}^{5}+2{x}^{3}+3x+1 \ g(f(x))=x.\, \Rightarrow {g}^{'}(f(x)){f}^{'}(x)=1 \end{gathered}$

$\begin{gathered} \frac{g(7)}{{g}^{'}(7)} \ g(7)\Rightarrow f(x)=7 \ {x}^{5}+2{x}^{3}+3x+1=7 \ \Rightarrow x({x}^{4}+2{x}^{2}+3)=0 \ \Rightarrow x=1 \ 	herefore g(7)\Rightarrow g(f(1))=1 \ {g}^{'}(f(x))=\frac{1}{{f}^{'}(x)} \ {g}^{'}(7) \ \Rightarrow f(x)=7\Rightarrow x=1 \ 	herefore {g}^{'}(7)=\frac{1}{{f}^{'}(1)} \ =\frac{1}{5{x}^{4}+6{x}^{2}+3} \ =\frac{1}{14} \ 	herefore \frac{g(7)}{{g}^{'}(7)}=\frac{1}{\frac{1}{14}}\, 14 \end{gathered}$

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