Question is:
Let $ f(x)=(1/2)x^3+6x+4 $ and let $y=f^{-1}(x)$ be the inverse function of f. Determine the x-coordinates of the two points on the graph of the inverse function where the tangent line is perpendicular to the straight line $y=-12x-9$.
Since (I believe) the inverse of the function is much too complicated to calculate easily, I figure the solution will involve the use of this formula: $\frac{d}{dx} f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$, but I have not figured out how.
Also, I think it is pretty obvious that the slope of the tangent will be $1/12$