Question

Please help me solve part "b)" of this question, I think I already know "a)"

Answer 1

`x=0,1`

Step-by-step explanation:

We were given that:

`\begin{gathered} y=(4x^3)/(6x-4) \\ y^(\prime)=(12x^2(x-1))/((3x-2)^2) \end{gathered}`

b)

The horizontal tangent line refers to where a function's derivative is zero since horizontal lines have a slope of zero.

When the function has horizontal tangent lines, we have:

`(12x^(2)(x-1))/((3x-2)^(2))=0`

Let's proceed to solve, we have:

`\begin{gathered} (12x^(2)(x-1))/((3x-2)^(2))=0 \\ \text{Cross multiply, we have:} \\ 12x^2(x-1)=0 \\ \text{Equating to zero, we have:} \\ 12x^2=0,x-1=0 \\ x^2=0,x=1 \\ x=√(0),x=1 \\ x=0,x=1 \end{gathered}`

Therefore, x is equal to: 0, 1