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If $f(x)=x^2-x+5, x>\frac{1}{2}$, and $\mathrm{g}(\mathrm{x})$ is its inverse function, then $\mathrm{g}^{\prime}(7)$ equals:
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Verified Answer
The correct answer is:
$\frac{1}{3}$
$\frac{1}{3}$
$f(x)=y=x^2-x+5$
$$
\begin{aligned}
&x^2-x+\frac{1}{4}-\frac{1}{4}+5=y \\
&\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=y \\
&\left(x-\frac{1}{2}\right)^2=y-\frac{19}{4} \\
&x-\frac{1}{2}=\pm \sqrt{y-\frac{19}{4}}
\end{aligned}
$$
$$
\begin{aligned}
&x=\frac{1}{2} \pm \sqrt{y-\frac{19}{4}} \\
&\text { As } x>\frac{1}{2} \\
&x=\frac{1}{2}+\sqrt{y-\frac{19}{4}} \\
&g(x)=\frac{1}{2}+\sqrt{x-\frac{19}{4}} \\
&g^{\prime}(x)=\frac{1}{2 \sqrt{x-\frac{19}{4}}}
\end{aligned}
$$
$$
g^{\prime}(7)=\frac{1}{2 \sqrt{7-\frac{19}{4}}}=\frac{1}{2 \frac{\sqrt{28-19}}{2}}=\frac{1}{3}
$$
$$
\begin{aligned}
&x^2-x+\frac{1}{4}-\frac{1}{4}+5=y \\
&\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=y \\
&\left(x-\frac{1}{2}\right)^2=y-\frac{19}{4} \\
&x-\frac{1}{2}=\pm \sqrt{y-\frac{19}{4}}
\end{aligned}
$$
$$
\begin{aligned}
&x=\frac{1}{2} \pm \sqrt{y-\frac{19}{4}} \\
&\text { As } x>\frac{1}{2} \\
&x=\frac{1}{2}+\sqrt{y-\frac{19}{4}} \\
&g(x)=\frac{1}{2}+\sqrt{x-\frac{19}{4}} \\
&g^{\prime}(x)=\frac{1}{2 \sqrt{x-\frac{19}{4}}}
\end{aligned}
$$
$$
g^{\prime}(7)=\frac{1}{2 \sqrt{7-\frac{19}{4}}}=\frac{1}{2 \frac{\sqrt{28-19}}{2}}=\frac{1}{3}
$$
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