Answer:
(1/2, 6).
Step-by-step explanation:
Turning points in a graph may exist whenever the derivative equals 0. Hence, let’s differentiate and then solve our function.
We have:
[tex]\displaystyle y=8x^2+\frac{2}{x}[/tex]
Take the derivative of both sides with respect to x:
[tex]\displaystyle y^\prime=16x-\frac{2}{x^2}[/tex]
Simplify:
[tex]\displaystyle y^\prime=\frac{16x^3-2}{x^2}[/tex]
We will now set this equal to 0 and solve for x. Hence:
[tex]\displaystyle \begin{aligned} 0&=\frac{16x^3-2}{x^2} \\ 0&=16x^3-2 \\ 2&=16x^3 \\ \frac{1}{8}&=x^3 \\ \frac{1}{2}&=x \end{aligned}[/tex]
Hence, the graph turns at x=1/2. This is the x-coordinate.
Then it follows that the y-coordinate will be:
[tex]\displaystyle{\begin{aligned} y&=8(\frac{1}{2})^2+\frac{2}{\frac{1}{2}} \\ y&=8(1/4)+2(2) \\ y&=2+4 \\ y&=6\end{aligned}[/tex]
Hence, our coordinate Is (1/2, 6).
