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

Question

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

Accepted Answer

$$\begin{aligned}
& f(x)=x^5+2 x^3+3 x+1 \
& g(f(x))=x . \quad \Rightarrow g^{\prime}(f(x)) f^{\prime}(x)=1 \
\end{aligned}$$

$$\begin{aligned}
&\begin{aligned}
& \text { Now } \frac{g(7)}{g^{\prime}(7)} \
& g(7) \Rightarrow f(x)=7 \
& x^5+2 x^3+3 x+1=7 \
& \Rightarrow x\left(x^4+2 x^2+3\right)=0 \
& \Rightarrow x=1 \
& \therefore g(7) \Rightarrow g(f(1))=1 \
& & ~g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} \
& g^{\prime}(7) \
& \Rightarrow f(x)=7 \Rightarrow x=1 \
& \therefore g^{\prime}(7)=\frac{1}{f^{\prime}(1)} \
& =\frac{1}{5 x^4+6 x^2+3} \
& =\frac{1}{14} \
& \therefore \frac{g(7)}{g^{\prime}(7)}=\frac{1}{\frac{1}{14}} \quad 14
\end{aligned}\
\end{aligned}$$

Solutions · 1