Question

The curve with the equation y=8x^2 + 2/x has one turning point.

Answer 1

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:

$\displaystyle y=8x^2+\frac{2}{x}$

Take the derivative of both sides with respect to x:

$\displaystyle y^\prime=16x-\frac{2}{x^2}$

Simplify:

$\displaystyle y^\prime=\frac{16x^3-2}{x^2}$

We will now set this equal to 0 and solve for x. Hence:

$\displaystyle \begin{aligned} 0&=\frac{16x^3-2}{x^2} \\ 0&=16x^3-2 \\ 2&=16x^3 \\ \frac{1}{8}&=x^3 \\ \frac{1}{2}&=x \end{aligned}$

Hence, the graph turns at x=1/2. This is the x-coordinate.

Then it follows that the y-coordinate will be:

$\displaystyle{\begin{aligned} y&=8(\frac{1}{2})^2+\frac{2}{\frac{1}{2}} \\ y&=8(1/4)+2(2) \\ y&=2+4 \\ y&=6\end{aligned}$

Hence, our coordinate Is (1/2, 6).