1. **State the problem:** We are given the curve equation $$y = k(3x - k)^{-1} + 3x$$ where $k$ is a constant. We need to find the values of $x$ at which the curve has stationary points, i.e., where the derivative $\frac{dy}{dx} = 0$. 2. **Recall the formula:** Stationary points occur where the first derivative of $y$ with respect to $x$ is zero: $$\frac{dy}{dx} = 0$$ 3. **Differentiate the function:** Given $$y = k(3x - k)^{-1} + 3x$$ Use the chain rule for the first term: $$\frac{d}{dx} \left[k(3x - k)^{-1}\right] = k \cdot (-1)(3x - k)^{-2} \cdot 3 = -3k(3x - k)^{-2}$$ The derivative of the second term is: $$\frac{d}{dx}(3x) = 3$$ So, $$\frac{dy}{dx} = -3k(3x - k)^{-2} + 3$$ 4. **Set the derivative equal to zero to find stationary points:** $$-3k(3x - k)^{-2} + 3 = 0$$ 5. **Solve for $x$:** $$-3k(3x - k)^{-2} = -3$$ Divide both sides by $-3$: $$k(3x - k)^{-2} = 1$$ Rewrite: $$\frac{k}{(3x - k)^2} = 1$$ Multiply both sides by $(3x - k)^2$: $$k = (3x - k)^2$$ Take square root of both sides: $$3x - k = \pm \sqrt{k}$$ 6. **Express $x$ in terms of $k$:** $$3x = k \pm \sqrt{k}$$ $$x = \frac{k \pm \sqrt{k}}{3}$$ **Final answer:** The stationary points occur at $$x = \frac{k + \sqrt{k}}{3} \quad \text{and} \quad x = \frac{k - \sqrt{k}}{3}$$