find all x - coordinates of points (x,y) on the curve ( y=\frac{(x - 7)^{5}}{(x - 3)^{3}} ) where the…

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}).

[ \begin{align*} y^\prime&=\frac{5(x - 7)^{4}(x - 3)^{3}-(x - 7)^{5}\cdot3(x - 3)^{2}}{(x - 3)^{6}}\ &=\frac{(x - 7)^{4}(x - 3)^{2}[5(x - 3)-3(x - 7)]}{(x - 3)^{6}}\ &=\frac{(x - 7)^{4}(5x-15 - 3x + 21)}{(x - 3)^{4}}\ &=\frac{(x - 7)^{4}(2x + 6)}{(x - 3)^{4}} \end{align*} ]

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\neq3)).

((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)