1. **State the problem:** We have a curve defined by the function $$y = (x + 1)(2x - 3)^4$$ and want to find the approximate change in $$y$$ as $$x$$ increases from 2 to $$2 + p$$, where $$p$$ is small. 2. **Formula and rule:** The approximate change in $$y$$ for a small change in $$x$$ can be found using the derivative: $$\Delta y \approx y'(x) \cdot \Delta x$$ where $$y'(x)$$ is the derivative of $$y$$ with respect to $$x$$ and $$\Delta x = p$$. 3. **Find the derivative:** Use the product rule for $$y = u \cdot v$$ where $$u = (x+1)$$ and $$v = (2x-3)^4$$. - Derivative of $$u$$: $$u' = 1$$ - Derivative of $$v$$: $$v' = 4(2x-3)^3 \cdot 2 = 8(2x-3)^3$$ (chain rule) Using product rule: $$y' = u'v + uv' = (1)(2x-3)^4 + (x+1) \cdot 8(2x-3)^3$$ 4. **Simplify the derivative:** $$y' = (2x-3)^3 \left[(2x-3) + 8(x+1)\right] = (2x-3)^3 (2x - 3 + 8x + 8) = (2x-3)^3 (10x + 5)$$ 5. **Evaluate the derivative at $$x=2$$:** - Calculate $$2x - 3 = 2(2) - 3 = 4 - 3 = 1$$ - Calculate $$10x + 5 = 10(2) + 5 = 20 + 5 = 25$$ So, $$y'(2) = 1^3 \times 25 = 25$$ 6. **Approximate change in $$y$$:** $$\Delta y \approx y'(2) \cdot p = 25p$$ **Final answer:** The approximate change in $$y$$ as $$x$$ increases from 2 to $$2 + p$$ is $$25p$$, where $$p$$ is small.