Question

$f\left(x\right)=x^2-x+5$ and $g\left(x\right)=f^{-1}\left(x\right)$. Then $g^{\prime }\left(7\right)=$

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{13}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{1}{13}$

$f\left(x\right)=x^2-x+5=y$
$g\left(x\right)=f^{-1}\left(x\right)$
$y=x^2-x+5$
$y=x^2-x+\frac{1}{4}-\frac{1}{4}+5$
$y={(x-\frac{1}{2})}^2+\frac{19}{4}$

$\sqrt{y-\frac{19}{4}}+\frac{1}{2}=x$
$\therefore g\left(x\right)=\sqrt{y-\frac{19}{4}}+\frac{1}{2}$
$g^{\prime }\left(x\right)=\frac{1}{2\sqrt{y-\frac{19}{4}}}$
when, $x=7,y=47$
$g^{\prime }\left(7\right)=\frac{1}{2\sqrt{47-\frac{19}{4}}}$
$=\frac{1}{\sqrt{188-19}}$
$=\frac{1}{\sqrt{169}}$
$g^{\prime }\left(7\right)=\frac{1}{13}$