1. **State the problem:** We are given the function $$f(x) = 4(3x - 4)^{-1} + 3x$$ for $$x \geq \frac{3}{2}$$ and need to find the stationary value at $$x = a$$, i.e., find $$a$$ where $$f'(a) = 0$$, and determine the nature of this stationary value. 2. **Find the derivative:** Use the derivative rules. For $$f(x) = 4(3x - 4)^{-1} + 3x$$, $$f'(x) = 4 \cdot \frac{d}{dx} (3x - 4)^{-1} + \frac{d}{dx} (3x)$$ Recall $$\frac{d}{dx} (u^{-1}) = -u^{-2} \cdot u'$$ where $$u = 3x - 4$$ and $$u' = 3$$. So, $$f'(x) = 4 \cdot (-1)(3x - 4)^{-2} \cdot 3 + 3 = -12(3x - 4)^{-2} + 3$$ 3. **Set derivative to zero to find stationary points:** $$-12(3x - 4)^{-2} + 3 = 0$$ Rearranged: $$3 = 12(3x - 4)^{-2}$$ Divide both sides by 3: $$1 = 4(3x - 4)^{-2}$$ Invert both sides: $$(3x - 4)^2 = 4$$ 4. **Solve for $$x$$:** $$3x - 4 = \pm 2$$ Case 1: $$3x - 4 = 2 \Rightarrow 3x = 6 \Rightarrow x = 2$$ Case 2: $$3x - 4 = -2 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3}$$ Since the domain is $$x \geq \frac{3}{2}$$, only $$x = 2$$ is valid. 5. **Determine the nature of the stationary point:** Find the second derivative: $$f''(x) = \frac{d}{dx} \left(-12(3x - 4)^{-2} + 3\right) = -12 \cdot (-2)(3x - 4)^{-3} \cdot 3 + 0 = 72(3x - 4)^{-3}$$ Evaluate at $$x = 2$$: $$f''(2) = 72(3(2) - 4)^{-3} = 72(6 - 4)^{-3} = 72(2)^{-3} = 72 \cdot \frac{1}{8} = 9 > 0$$ Since $$f''(2) > 0$$, the stationary point at $$x = 2$$ is a local minimum. **Final answer:** $$a = 2$$ and the stationary value is a local minimum.