Question

If $f\left(x\right)=x^2-x+5,x>\frac{1}{2}$, and $g\left(x\right)$ is its inverse function, then $g^{\prime }\left(7\right)$ equals :

• A. $-\frac{1}{3}$
• B. $\frac{1}{13}$
• C. $\frac{1}{3}$
• D. $-\frac{1}{13}$

Answer 1

Answer: (C)

$\frac{1}{3}$

$f\left(x\right)=x^2-x+5,x>\frac{1}{2}$

$y=x^2-x+5$

$x^2-x+5-y=0$

$x=\frac{1\pm \sqrt{1+4(y-5)}}{2}$

$g\left(x\right)=\frac{1\pm \sqrt{1+4(x-5)}}{2}$

$g^{\prime }\left(x\right)=4\times \frac{1}{2\sqrt{1+4(x-5)}}\times \frac{1}{2}$

$g^{\prime }\left(7\right)=\frac{1}{3}$