Solution — 2024 Question 3 Graphs| Differentiation & Applications

(a)

\begin{matrix} y & = 4 x^{2} + \frac{27}{x} - 35 \\ \frac{d y}{d x} & = 8 x + 27 (- x^{- 2}) \\ = 8 x - 27 x^{- 2} \\ = 8 x - \frac{27}{x^{2}} \end{matrix}

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

\begin{matrix} 8 x - \frac{27}{x^{2}} & = 0 \\ 8 x & = \frac{27}{x^{2}} \\ 8 x^{3} & = 27 \\ x^{3} & = \frac{27}{8} \\ x & = \frac{3}{2} \end{matrix}

When $x = \frac{3}{2},$

\begin{matrix} y & = 4 {(\frac{3}{2})}^{2} + \frac{27}{\frac{3}{2}} - 35 \\ = - 8 \end{matrix}

Coordinates of the stationary point on $C$ : $(\frac{3}{2}, - 8) ∎$

(b)