Solution: –

(a) Find \(\frac{dy}{dx}\) and show that the x-coordinate of \(P\) satisfies \(x = \frac{1}{6} + \frac{1}{2}e^{-2x}\)

The curve is \(y = \frac{1 + e^{2x}}{1 + 3x}\). To find the stationary point \(P\), we need \(\frac{dy}{dx} = 0\).

Use the quotient rule:
\[
\frac{dy}{dx} = \frac{\frac{d}{dx}(1 + e^{2x}) \cdot (1 + 3x) – (1 + e^{2x}) \cdot \frac{d}{dx}(1 + 3x)}{(1 + 3x)^2}.
\]
Derivative of the numerator \(1 + e^{2x}\): \(\frac{d}{dx}(1 + e^{2x}) = 0 + 2e^{2x} = 2e^{2x}\).
Derivative of the denominator \(1 + 3x\): \(\frac{d}{dx}(1 + 3x) = 3\).

So,
\[
\frac{dy}{dx} = \frac{(2e^{2x})(1 + 3x) – (1 + e^{2x}) \cdot 3}{(1 + 3x)^2}.
\]
Simplify the numerator:
\[
2e^{2x}(1 + 3x) – 3(1 + e^{2x}) = 2e^{2x} + 6x e^{2x} – 3 – 3e^{2x} = (2e^{2x} – 3e^{2x}) + 6x e^{2x} – 3 = -e^{2x} + 6x e^{2x} – 3.
\]
\[
\frac{dy}{dx} = \frac{-e^{2x} + 6x e^{2x} – 3}{(1 + 3x)^2}.
\]
Set \(\frac{dy}{dx} = 0\) for the stationary point:
\[
e^{2x} + 6x e^{2x} – 3 = 0.
\]
Factor out \(e^{2x}\) (since \(e^{2x} \neq 0\)):
\[
e^{2x}(6x – 1) – 3 = 0.
\]
\[
e^{2x}(6x – 1) = 3.
\]
\[
6x – 1 = 3e^{-2x}.
\]
\[
6x = 1 + 3e^{-2x}.
\]
\[
x = \frac{1}{6} + \frac{3}{6}e^{-2x} = \frac{1}{6} + \frac{1}{2}e^{-2x}.
\]
This matches the given equation, so the x-coordinate of \(P\) satisfies \(x = \frac{1}{6} + \frac{1}{2}e^{-2x}\).

(b) Show that the x-coordinate of \(P\) lies between 0.35 and 0.45

We need to check if there’s a solution to \(x = \frac{1}{6} + \frac{1}{2}e^{-2x}\) in the interval \(0.35 < x < 0.45\).

Define the function:
\[
f(x) = x – \frac{1}{6} – \frac{1}{2}e^{-2x}.
\]
We need to show \(f(x) = 0\) for some \(x\) in \((0.35, 0.45)\).

Evaluate \(f(x)\) at the endpoints and check the behavior:
At \(x = 0.35\):
\[
e^{-2 \cdot 0.35} = e^{-0.7} \approx 0.4966.
\]
\[
\frac{1}{2} \cdot 0.4966 \approx 0.2483.
\]
\[
\frac{1}{6} \approx 0.1667.
\]
\[
f(0.35) = 0.35 – 0.1667 – 0.2483 = 0.35 – 0.415 = -0.065.
\]
At \(x = 0.45\):
\[
e^{-2 \cdot 0.45} = e^{-0.9} \approx 0.4066.
\]
\[
\frac{1}{2} \cdot 0.4066 \approx 0.2033.
\]
\[
f(0.45) = 0.45 – 0.1667 – 0.2033 = 0.45 – 0.37 = 0.08.
\]
\(f(0.35) < 0\) and \(f(0.45) > 0\), so by the Intermediate Value Theorem, there’s a root in \((0.35, 0.45)\).

To confirm it’s the only root (since the curve has exactly one stationary point), check the derivative of \(f(x)\):
\[
f'(x) = 1 – \frac{1}{2} \cdot (-2) e^{-2x} = 1 + e^{-2x}.
\]
Since \(e^{-2x} > 0\), \(f'(x) > 1 > 0\), so \(f(x)\) is strictly increasing. Thus, there’s exactly one root, and it lies in \((0.35, 0.45)\).

(c) Use an iterative formula to find the x-coordinate of \(P\) correct to 3 significant figures, giving each iteration to 5 significant figures

From (a), \(x = \frac{1}{6} + \frac{1}{2}e^{-2x}\). Rearrange for iteration:
\[
x_{n+1} = \frac{1}{6} + \frac{1}{2}e^{-2x_n}.
\]
Start with an initial guess within \((0.35, 0.45)\), say \(x_0 = 0.4\) (midpoint for convergence).

\(x_0 = 0.40000\):
\[
e^{-2 \cdot 0.40000} = e^{-0.8} \approx 0.44933.
\]
\[
\frac{1}{2} \cdot 0.44933 = 0.22467.
\]
\[
x_1 = 0.16667 + 0.22467 = 0.39134.
\]
\(x_1 = 0.39134\):
\[
e^{-2 \cdot 0.39134} = e^{-0.78268} \approx 0.45726.
\]
\[
\frac{1}{2} \cdot 0.45726 = 0.22863.
\]
\[
x_2 = 0.16667 + 0.22863 = 0.39530.
\]
\(x_2 = 0.39530\):
\[
e^{-2 \cdot 0.39530} = e^{-0.79060} \approx 0.45370.
\]
\[
\frac{1}{2} \cdot 0.45370 = 0.22685.
\]
\[
x_3 = 0.16667 + 0.22685 = 0.39352.
\]
\(x_3 = 0.39352\):
\[
e^{-2 \cdot 0.39352} = e^{-0.78704} \approx 0.45515.
\]
\[
\frac{1}{2} \cdot 0.45515 = 0.22758.
\]
\[
x_4 = 0.16667 + 0.22758 = 0.39425.
\]
\(x_4 = 0.39425\):
\[
e^{-2 \cdot 0.39425} = e^{-0.78850} \approx 0.45437.
\]
\[
\frac{1}{2} \cdot 0.45437 = 0.22719.
\]
\[
x_5 = 0.16667 + 0.22719 = 0.39386.
\]
\(x_5 = 0.39386\):
\[
e^{-2 \cdot 0.39386} = e^{-0.78772} \approx 0.45482.
\]
\[
\frac{1}{2} \cdot 0.45482 = 0.22741.
\]
\[
x_6 = 0.16667 + 0.22741 = 0.39408.
\]
\(x_6 = 0.39408\):
\[
e^{-2 \cdot 0.39408} = e^{-0.78816} \approx 0.45452.
\]
\[
\frac{1}{2} \cdot 0.45452 = 0.22726.
\]
\[
x_7 = 0.16667 + 0.22726 = 0.39393.
\]

The values are converging. To 3 significant figures:
\(x_6 = 0.39408\), \(x_7 = 0.39393\), both round to 0.394.
So, the x-coordinate of \(P\) is 0.394 (to 3 significant figures).

Iterations to 5 significant figures:
\(x_0 = 0.40000\)
\(x_1 = 0.39134\)
\(x_2 = 0.39530\)
\(x_3 = 0.39352\)
\(x_4 = 0.39425\)
\(x_5 = 0.39386\)
\(x_6 = 0.39408\)
\(x_7 = 0.39393\)