Question

find all X-coordinates of points (x,y) on the curve y=(x-5)^6/(x-4)^5 where the tangent line is horizontal

Answer 1

Given:

`y=((x-5)^6)/((x-4)^5)`

To Determine: Where the tangent line is horizontal

Solution

Please note where the tangent line is horizontal is when the derivative is equal to zero

Determine the derivative of the function using quotient rule

`\begin{gathered} Quotient\text{ rule} \\ If,y=(u)/(v),then,(dy)/(dx)=(v(du)/(dx)-u(dv)/(dx))/(v^2) \end{gathered}`
`\begin{gathered} Given \\ y=((x-5)^6)/((x-4)^5) \\ u=(x-5)^6 \\ v=(x-4)^5 \end{gathered}`
`\begin{gathered} u=(x-5)^6 \\ (du)/(dx)=6(x-5)^5 \\ v=(x-4)^5 \\ (dv)/(dx)=5(x-4)^4 \end{gathered}`
`\begin{gathered} Therefore \\ (dy)/(dx)=((x-4)^5*6(x-5)^5-(x-5)^6*5(x-4)^4)/(((x-4)^5)^2) \end{gathered}`
`(dy)/(dx)=((x-4)^4(x-5)^5((6(x-4)-5(x-5)))/((x-4)^(10))`
`\begin{gathered} (dy)/(dx)=(\left(x-4\right)^4\left(x-5\right)^5\left(x+1\right))/(\left(x-4\right)^(10)) \\ (dy)/(dx)=(\left(x-4\right)^4\left(x-5\right)^5\left(x+1\right))/(\left(x-4\right)^4\left(x-4\right)^6) \\ (dy)/(dx)=(\left(x-5\right)^5\left(x+1\right))/(\left(x-4\right)^6) \end{gathered}`

Equate the derivative to zero

`\begin{gathered} (\left(x-5\right)^5\left(x+1\right))/(\left(x-4\right)^6)=0 \\ \left(x-5\right)^5\left(x+1\right)=0 \\ (x-5)^5=0,or,x+1=0 \\ x-5=0,or,x+1=0 \\ x=5,or,x=-1 \end{gathered}`
`\begin{gathered} when,x=5 \\ y=((5-5)^6)/((5-4)^5)=(0^6)/(1^5)=(0)/(1)=0 \end{gathered}`
`\begin{gathered} When,x=-1 \\ y=((-1-5)^6)/((-1-4)^5)=((-6)^6)/((-5)^5)=(46656)/(-3125)=-14.92992 \end{gathered}`

Hence

(5,0) and (-1,-14.92992)

x=(-1, 5)