To find the coordinates of the stationary point on the curve $y = 8x^2 + \frac{2}{x}$, we need to determine the points where the derivative $\frac{dy}{dx}$ is zero.

Step 1: Differentiate the equation

The given equation is: $y = 8x^2 + \frac{2}{x}$

Let's find $\frac{dy}{dx}$:

1. Differentiate $8x^2$ with respect to $x$: $\frac{d}{dx}(8x^2) = 16x$

2. Differentiate $\frac{2}{x}$ with respect to $x$: $\frac{d}{dx}\left(\frac{2}{x}\right) = -\frac{2}{x^2}$

So, $\frac{dy}{dx} = 16x - \frac{2}{x^2}$

Step 2: Set the derivative to zero to find stationary points

To find the stationary points, set $\frac{dy}{dx} = 0$: $16x - \frac{2}{x^2} = 0$

Multiply both sides by $x^2$ to eliminate the fraction: $16x^3 - 2 = 0$

Rearrange to solve for $x$: $16x^3 = 2$ $x^3 = \frac{2}{16}$ $x^3 = \frac{1}{8}$

Taking the cube root of both sides: $x = \frac{1}{2}$

Step 3: Substitute $x = \frac{1}{2}$ back into the original equation to find $y$

Now that we know $x = \frac{1}{2}$, substitute it into the original equation to find $y$: $y = 8\left(\frac{1}{2}\right)^2 + \frac{2}{\frac{1}{2}}$

Calculate each term separately:

1. $8\left(\frac{1}{2}\right)^2 = 8 \cdot \frac{1}{4} = 2$
2. $\frac{2}{\frac{1}{2}} = 2 \times 2 = 4$

So, $y = 2 + 4 = 6$

Conclusion

The coordinates of the stationary point are: $\left( \frac{1}{2}, 6 \right)$

Would you like further explanation or have any questions?

---

Here are some related questions:

1. How do we determine the nature of the stationary point?
2. What happens if we differentiate the function again to find concavity?
3. Could we use a graphical approach to verify the stationary point?
4. What are the implications of stationary points in real-world applications?
5. How can we find other features of the curve, like asymptotes?

Tip: Checking the second derivative can help you determine whether a stationary point is a maximum, minimum, or a point of inflection.