Let $f(x) = \frac14x^3 + 12x + 6$ 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 $y = -24x – 32$.
Need help on how to do this, if anyone could show me it would be highly appreciated.
I calculated the derivative to be $\frac34x^2 + 12$. From here do I find the inverse of the derivative and the inverse of the straight line and set them equal to each other to find the $x$-coordinates, or am I supposed to do something else to solve this?
Thanks
Best Answer
$y = \frac 1 4 x^3 + 12x + 6$
$L = -24x - 32$
$P$ is a line that is perpendicular to $L$.
The slope of a perpendicular is the negative of the multiplicative inverse, that is: $$\frac{dP} {dx} = -(\frac {dL} {dx})^{-1} = -\frac{dx} {dL}$$
To solve the problem we want $y$ where: $$\underbrace{\frac {dx} {dy}}_{\text{slope of the inverse}} = \underbrace{\frac {dP} {dx}}_{\text{Slope of a perpendicular}}$$ $$\frac {dx} {dy} = -\frac {dx} {dL}$$ $$\frac {dy} {dx} = -\frac {dL} {dx}$$ $$\frac {3} {4} x^2 + 12 = -(-24)$$ $$x = \pm 4$$
Then just find the corresponding $y$ values, which are the $x$ values of the inverse function.