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We are given the curve with the equation:

\[
y = 4x - \frac{1}{2}e^{2x}
\]

We are tasked to:
- **(a)** Find the coordinates of the stationary point of the curve in terms of natural logarithms.
- **(b)** Determine the nature of the stationary point.

### (a) Finding the stationary point
The stationary point occurs where the derivative of \(y\) with respect to \(x\) is equal to zero. First, let's differentiate \(y\).

\[
\frac{dy}{dx} = \frac{d}{dx} \left( 4x - \frac{1}{2}e^{2x} \right)
\]

Using the linearity of differentiation and applying the chain rule for the exponential term:

\[
\frac{dy}{dx} = 4 - \frac{1}{2} \cdot 2e^{2x} = 4 - e^{2x}
\]

Now, to find the stationary point, set \(\frac{dy}{dx} = 0\):

\[
4 - e^{2x} = 0
\]

Solving for \(x\):

\[
e^{2x} = 4
\]
\[
2x = \ln(4)
\]
\[
x = \frac{1}{2} \ln(4)
\]

Since \(\ln(4) = 2 \ln(2)\), we can simplify further:

\[
x = \ln(2)
\]

Now, substitute \(x = \ln(2)\) into the original equation for \(y\) to find the \(y\)-coordinate:

\[
y = 4(\ln(2)) - \frac{1}{2}e^{2(\ln(2))}
\]

Since \(e^{2 \ln(2)} = 4\), this simplifies to:

\[
y = 4\ln(2) - \frac{1}{2} \cdot 4 = 4\ln(2) - 2
\]

Thus, the coordinates of the stationary point are:

\[
\left( \ln(2), 4\ln(2) - 2 \right)
\]

### (b) Determining the nature of the stationary point

To determine the nature of the stationary point, we need to check the second derivative of \(y\).

The first derivative is:

\[
\frac{dy}{dx} = 4 - e^{2x}
\]

Now, differentiate this again to get the second derivative:

\[
\frac{d^2y}{dx^2} = -2e^{2x}
\]

At \(x = \ln(2)\), we substitute this into the second derivative:

\[
\frac{d^2y}{dx^2} = -2e^{2\ln(2)} = -2 \times 4 = -8
\]

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point is a **maximum**.

---

### Final Answer:
- **(a)** The coordinates of the stationary point are \(\left( \ln(2), 4\ln(2) - 2 \right)\).
- **(b)** The stationary point is a **maximum**.

Would you like further details or have any questions?

Here are 5 related questions:
1. What does the second derivative tell us about a function's concavity?
2. How does the chain rule apply when differentiating exponential functions?
3. Why is the natural logarithm used in solving for the stationary point?
4. Can a stationary point be neither a maximum nor a minimum? If so, how?
5. How would the behavior of the curve change if the coefficient of \(e^{2x}\) was different?

**Tip:** For determining the nature of stationary points, always check the second derivative. A positive value indicates a minimum, while a negative value indicates a maximum.

A curve has the equation y = 4x - (1/2)e^(2x). a) Find the coordinates of the stationary point of the curve, giving your answers in terms of natural logarithms. b) Determine the nature of the stationary point.

Learn how to find and classify the stationary point of the curve y = 4x - (1/2)e^(2x). This solution includes the use of differentiation and the second derivative test to determine the nature of the stationary point. Suitable for students in Grades 11-12.

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