Question

find all x - coordinates of points (x,y) on the curve ( y=\frac{(x - 7)^{5}}{(x - 3)^{3}} ) where the tangent line is horizontal.
( x=square )
(use a comma to separate answers as needed.)

Explanation:

Step1: Differentiate the function using the quotient rule

The quotient rule states that if \(y=\frac{u}{v}\), then \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here, \(u=(x - 7)^{5}\), so \(u^\prime = 5(x - 7)^{4}\) (using the chain rule \((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)\), with \(f(t)=t^{5}\), \(g(x)=x - 7\)), and \(v=(x - 3)^{3}\), so \(v^\prime=3(x - 3)^{2}\).

$$
LATEXBLOCK0
$$

Step2: Set the derivative equal to zero

A horizontal tangent line occurs when \(y^\prime = 0\). Since \(\frac{(x - 7)^{4}(2x + 6)}{(x - 3)^{4}}=0\), we set the numerator equal to zero (while considering the domain \(x
eq3\)).

\((x - 7)^{4}(2x + 6)=0\)

We have two cases:

Case 1: \((x - 7)^{4}=0\), then \(x = 7\)

Case 2: \(2x+6=0\), then \(2x=-6\), \(x=-3\)

Answer:

\(-3,7\)