Find all x-coordinates of points (x,y) on the curve of where the tangent line is horizontal. (Thanks!!)
The tangent line will be horizontal when y'(x) = 0. So:
$\displaystyle y(x) = \frac{(x-2)^5}{(x-4)^3}$
Using the quotient rule:
$\displaystyle y'(x) = \frac{5(x-2)^4(x-4)^3 - (x-2)^53(x-4)^2}{(x-4)^6}$
Doing a little factoring:
$\displaystyle y'(x) = \frac{(x-2)^4(x-4)^2}{(x-4)^6} \cdot (5(x-4) - 3(x-2))$
$\displaystyle y'(x) = \frac{(x-2)^4}{(x-4)^4} \cdot (5x - 20 - 3x + 6)$
$\displaystyle y'(x) = \frac{(x-2)^4}{(x-4)^4} \cdot (2x - 14)$
$\displaystyle y'(x) = \frac{2(x-2)^4(x - 7)}{(x-4)^4}$
We need to solve this for y'(x) = 0.
As long as $\displaystyle x \neq 4$ may simply set the numerator equal to 0:
$\displaystyle 0 = 2(x-2)^4(x - 7)$
This will obviously happen at x = 2 and x = 7.
-Dan