Question

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

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

Answer 1

Answer: (B)

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