the curve with equation y=8x^2 + 2/x find the co-ordinate of this stationary point. Show your working clearly.

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?

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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.

The curve with equation y=8x^2 + 2/x find the co-ordinate of this stationary point. Show your working clearly.

Learn how to find the stationary point of the curve y=8x^2 + 2/x. This step-by-step solution covers differentiation, solving for critical points, and finding the coordinate of the stationary point. Suitable for students in Grades 11-12.

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