Answer:
[tex]x=0,1[/tex]Explanation:
We were given that:
[tex]\begin{gathered} y=\frac{4x^3}{6x-4} \\ y^{\prime}=\frac{12x^2(x-1)}{(3x-2)^2} \end{gathered}[/tex]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:
[tex]\frac{12x^{2}(x-1)}{(3x-2)^{2}}=0[/tex]Let's proceed to solve, we have:
[tex]\begin{gathered} \frac{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=\sqrt{0},x=1 \\ x=0,x=1 \end{gathered}[/tex]Therefore, x is equal to: 0, 1