1. Stating the problem: We want to find the x-coordinate of the point on the curve $y = (x + 2)\sqrt{1 - 2x}$ where the gradient (derivative) is zero. 2. Express the function: Let $y = (x + 2)\sqrt{1 - 2x}$. 3. Rewrite $y$ to facilitate differentiation: Note that $\sqrt{1 - 2x} = (1 - 2x)^{\frac{1}{2}}$, so $$y = (x+2)(1 - 2x)^{\frac{1}{2}}.$$ 4. Use the product rule to find $\frac{dy}{dx}$: If $y = u v$ with $$u = x+2 \quad \text{and} \quad v = (1-2x)^{\frac{1}{2}},$$ then $$\frac{dy}{dx} = u'v + uv'.$$ 5. Compute the derivatives: $$u' = \frac{d}{dx}(x+2) = 1,$$ $$v' = \frac{d}{dx}(1-2x)^{\frac{1}{2}} = \frac{1}{2}(1-2x)^{-\frac{1}{2}} \cdot (-2) = -\frac{1}{(1-2x)^{\frac{1}{2}}}.$$ 6. Substitute back: $$\frac{dy}{dx} = 1 \cdot (1-2x)^{\frac{1}{2}} + (x+2) \cdot \left(-\frac{1}{(1-2x)^{\frac{1}{2}}}\right) = (1-2x)^{\frac{1}{2}} - \frac{x+2}{(1-2x)^{\frac{1}{2}}}.$$ 7. Combine into a single fraction: $$\frac{dy}{dx} = \frac{(1-2x) - (x+2)}{(1-2x)^{\frac{1}{2}}} = \frac{1 - 2x - x - 2}{(1-2x)^{\frac{1}{2}}} = \frac{-3x - 1}{(1-2x)^{\frac{1}{2}}}.$$ 8. Set the gradient equal to zero: $$\frac{dy}{dx} = 0 \implies \frac{-3x - 1}{(1-2x)^{\frac{1}{2}}} = 0.$$ Since the denominator cannot be zero (domain restriction: $1-2x > 0 \Rightarrow x < \frac{1}{2}$), the numerator must be zero: $$-3x -1 = 0 \, \Rightarrow \, 3x = -1 \, \Rightarrow \, x = -\frac{1}{3}.$$ 9. Check the domain: $x = -\frac{1}{3} < \frac{1}{2}$, so valid. **Final answer:** The x-coordinate where the gradient is zero is $$x = -\frac{1}{3}.$$