Suppose that $f(x)=\frac{2 x+1}{x-2}$ and that $y=g(x)$ is a linear function. If $f^{-1}(g(2))=7$ and $g^{-1}(f(1))=\frac{4}{5}$, what is the $x$-intercept of $y=g(x) ?$
Applying $f$ to both sides of $f^{-1}(g(2))=7$, we get

$$
f(7)=f\left(f^{-1}(g(2))\right)=g(2)
$$

so we have that $g(2)=f(7)=\frac{2(7)+1}{7-2}=\frac{15}{5}=3$.

Similarly, we can apply $g$ to both sides of $g^{-1}(f(1))=\frac{4}{5}$ to get

$$
g\left(\frac{4}{5}\right)=g\left(g^{-1}(f(1))\right)=f(1)
$$

so we have that $g\left(\frac{4}{5}\right)=f(1)=\frac{2(1)+1}{1-2}=-3$.

Since $g(2)=3$ and $g\left(\frac{4}{5}\right)=-3$, the $x$-intercept is halfway between $\frac{4}{5}$ and 2 .

The $x$-intercept is $\frac{1}{2}\left(2+\frac{4}{5}\right)=\frac{7}{5}$.

ANSWER: $\frac{7}{5}$