Question

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

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

Answer 1

Answer: (B)

$f(x)=x^2-x+5,x>\frac{1}{2}$
$y=x^2-x+5$
$x^2-x+5-y=0$
$x=\frac{1\pm \sqrt{1+4(y-5)}}{2}$
$g(x)=\frac{1\pm \sqrt{1+4(x-5)}}{2}$
$g^{\prime }(x)=4\times \frac{1}{2\sqrt{1}+4(x-5)}\times \frac{1}{2}$
$g^{\prime }(7)=\frac{1}{3}$