Solution — 2016 Question 4 Graphs| Differentiation & Applications| Integration & Applications

(a)

Differentiating with respect to $x$ :

\begin{matrix} y & = 1 + 6 x - 3 x^{2} - 4 x^{3} \\ \frac{d y}{d x} & = 6 - 6 x - 12 x^{2} ∎ \end{matrix}

At stationary points, $\frac{d y}{d x} = 0 :$

\begin{matrix} 6 - 6 x - 12 x^{2} & = 0 \\ 2 x^{2} + x - 1 & = 0 \\ (x + 1) (2 x - 1) & = 0 \end{matrix}

$x = - 1 or x = \frac{1}{2}$

When $x = - 1,$

\begin{matrix} y & = 1 + 6 (- 1) - 3 {(- 1)}^{2} - 4 {(- 1)}^{3} \\ = - 4 \end{matrix}

When $x = \frac{1}{2},$

\begin{matrix} y & = 1 + 6 (\frac{1}{2}) - 3 {(\frac{1}{2})}^{2} - 4 {(\frac{1}{2})}^{3} \\ = \frac{11}{4} \end{matrix}

Hence the coordinates of the turning points are

$(- 1, - 4) ∎ and (\frac{1}{2}, \frac{11}{4}) ∎$

(b)

\begin{matrix} \frac{d y}{d x} & = 6 - 6 x - 12 x^{2} \\ \frac{d^{2} y}{d x^{2}} & = - 6 - 24 x \end{matrix}

When $x = - 1,$

\begin{matrix} \frac{d^{2} y}{d x^{2}} & = - 6 - 24 (- 1) \\ = 18 \end{matrix}

Since $\frac{d^{2} y}{d x^{2}} > 0,$ the stationary point at $x = - 1$ is a minimum point $∎$

When $x = \frac{1}{2},$

\begin{matrix} \frac{d^{2} y}{d x^{2}} & = - 6 - 24 (- 1) \\ = 18 \end{matrix}

Since $\frac{d^{2} y}{d x^{2}} < 0,$ the stationary point at $x = \frac{1}{2}$ is a maximum point $∎$

(c)

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

The $x$ -intercepts are $(- 1.59, 0),$ $(- 0.157, 0)$ and $(1, 0)$ (3 s.f.). $∎$

(d)

\begin{matrix} Area \\ = \int_{0.5}^{1} (1 + 6 x - 3 x^{2} - 4 x^{3}) d x \\ = 0.937 50 \\ = 0.9375 ∎ {units}^{2} \end{matrix}