Curve Derivatives D86Dd7

1. **State the problem:** We are given the curve $$y = 3x + 1 - 4(3x + 1)^{1/2}$$ for $$x > -\frac{1}{3}$$. We need to find: (a) The first derivative $$\frac{dy}{dx}$$ and the second derivative $$\frac{d^2y}{dx^2}$$. (b) The coordinates of the stationary point and determine whether it is a maximum, minimum, or point of inflection. 2. **Find the first derivative $$\frac{dy}{dx}$$:** Recall the derivative rules: - Derivative of $$3x + 1$$ is $$3$$. - Derivative of $$-4(3x + 1)^{1/2}$$ uses the chain rule: $$\frac{d}{dx} (3x + 1)^{1/2} = \frac{1}{2}(3x + 1)^{-1/2} \times 3 = \frac{3}{2(3x + 1)^{1/2}}$$. So, $$\frac{dy}{dx} = 3 - 4 \times \frac{3}{2(3x + 1)^{1/2}} = 3 - \frac{12}{2(3x + 1)^{1/2}} = 3 - \frac{6}{(3x + 1)^{1/2}}$$ 3. **Find the second derivative $$\frac{d^2y}{dx^2}$$:** Differentiate $$\frac{dy}{dx} = 3 - 6(3x + 1)^{-1/2}$$: $$\frac{d^2y}{dx^2} = 0 - 6 \times \left(-\frac{1}{2}\right)(3x + 1)^{-3/2} \times 3 = 6 \times \frac{1}{2} \times 3 (3x + 1)^{-3/2} = 9(3x + 1)^{-3/2}$$ 4. **Find stationary points:** Stationary points occur where $$\frac{dy}{dx} = 0$$: $$3 - \frac{6}{(3x + 1)^{1/2}} = 0$$ Rearranged: $$\frac{6}{(3x + 1)^{1/2}} = 3$$ Multiply both sides by $$(3x + 1)^{1/2}$$: $$6 = 3(3x + 1)^{1/2}$$ Divide both sides by 3: $$2 = (3x + 1)^{1/2}$$ Square both sides: $$4 = 3x + 1$$ Solve for $$x$$: $$3x = 3 \Rightarrow x = 1$$ 5. **Find the corresponding $$y$$ coordinate:** Substitute $$x = 1$$ into the original equation: $$y = 3(1) + 1 - 4(3(1) + 1)^{1/2} = 3 + 1 - 4(4)^{1/2} = 4 - 4 \times 2 = 4 - 8 = -4$$ So the stationary point is at $$(1, -4)$$. 6. **Determine the nature of the stationary point:** Evaluate $$\frac{d^2y}{dx^2}$$ at $$x = 1$$: $$\frac{d^2y}{dx^2} = 9(3(1) + 1)^{-3/2} = 9(4)^{-3/2}$$ Calculate $$4^{-3/2}$$: $$4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$$ So, $$4^{-3/2} = \frac{1}{8}$$ Therefore, $$\frac{d^2y}{dx^2} = 9 \times \frac{1}{8} = \frac{9}{8} > 0$$ Since the second derivative is positive, the stationary point is a **local minimum**. **Final answers:** (a) $$\frac{dy}{dx} = 3 - \frac{6}{(3x + 1)^{1/2}}$$ $$\frac{d^2y}{dx^2} = 9(3x + 1)^{-3/2}$$ (b) Stationary point at $$(1, -4)$$ which is a local minimum.