Answer :
Given:
[tex]\begin{gathered} f(3)=-2 \\ f^{^{\prime}}(3)=-8 \\ g(x)=\frac{1}{f(x)} \end{gathered}[/tex]Solution:
If f(3)=-2, then f'(-2)=3
Hence, the points are (-2,3). These points will be on the graph of the inverse function to f(x).
The function g(x) is as
[tex]\begin{gathered} g(x)=y \\ =\frac{1}{f(x)} \end{gathered}[/tex]Now, the tangent line slop will be as
[tex]\begin{gathered} f(-2)=\frac{1}{f^{^{\prime}}(3)} \\ =\frac{1}{-8} \end{gathered}[/tex]The equation of the tangent line will be as
[tex]\begin{gathered} y-3=\frac{1}{-8}(x-(-2)) \\ y-3=\frac{1}{-8}(x+2) \end{gathered}[/tex]Answer:
Hence, the tangent line equation is as
[tex]y=\frac{1}{-8}(x+2)+3[/tex]