Solution — 2021 Question 3 Graphs| Differentiation & Applications

(a)

\begin{matrix} y & = x^{2} - 14 + \frac{16}{x} \\ \frac{d y}{d x} & = 2 x + 16 (- x^{- 2}) \\ = 2 x - 16 x^{- 2} \\ = 2 x - \frac{16}{x^{2}} \end{matrix}

At the stationary point, $\frac{d y}{d x} = 0$

\begin{matrix} 2 x - \frac{16}{x^{2}} & = 0 \\ 2 x & = \frac{16}{x^{2}} \\ 2 x^{3} & = 16 \\ x^{3} & = 8 \\ x & = 2 \end{matrix}

When $x = 2,$

\begin{matrix} y & = 2^{2} - 14 + \frac{16}{2} \\ = - 2 \end{matrix}

Hence the coordinates of the stationary point are $(2, - 2) ∎$

(b)

(c)

\begin{matrix} 2 x^{2} - 20 + \frac{16}{x} & = 0 \\ x^{2} - 14 + \frac{16}{x} & = - x^{2} + 6 \end{matrix}

The required curve is $y = - x^{2} + 6 ∎$

(d)

From GC,

$x = 0.865 ∎ or x = 2.64 ∎$