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If $f(3)=-2$, $f'(3)=-8$, and $g(x)$ is the inverse function to $f(x)$, the equation of the tangent line to $g(x)$ at $x=-2$ is:

$y=-\cfrac{1}{8} (x+2)+3$

$y=\cfrac{1}{8} (x+2)+3$

$y=8(x-3)-2$

$y=-8(x-3)-2$