Solution — 2020 Question 4 Graphs| Differentiation & Applications| Integration & Applications

(a)

Using a graphing calculator to solve $y = 0$ :

The $x$ -intercepts are $(- 0.732, 0),$ $(1.00, 0)$ and $(2.73, 0)$ (3 s.f.). $∎$

(b)

\begin{matrix} y & = x^{3} - 3 x^{2} + 2 \\ \frac{d y}{d x} & = 3 x^{2} - 6 x \end{matrix}

When $x = 0.5,$

\begin{matrix} y & = x^{3} - 3 x^{2} + 2 \\ = {0.5}^{3} - 3 {(0.5)}^{2} + 2 \\ = 1.375 \end{matrix}

\begin{matrix} \frac{d y}{d x} & = 3 {(0.5)}^{2} - 6 (0.5) \\ = - 2.25 \end{matrix}

Equation of tangent:

\begin{matrix} y - 1.375 & = - 2.25 (x - 0.5) \\ = - 2.25 x + 1.125 \\ y + 2.25 x & = 2.5 \\ 9 x + 4 y & = 10 ∎ \end{matrix}

(c)

\begin{matrix} Area \\ = \int_{3}^{4} (x^{3} - 3 x^{2} + 2) d x \\ = {[\frac{x^{4}}{4} - x^{3} + 2 x]}_{3}^{4} \\ = \frac{4^{4}}{4} - 4^{3} + 2 (4) - (\frac{3^{4}}{4} - 3^{3} + 2 (3)) \\ = \frac{35}{4} ∎ \end{matrix}