Solution — 2024 Question 3 Graphs| Differentiation & Applications

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

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

$$\begin{array}{cc}y& =4{\left(\frac{3}{2}\right)}^2+\frac{27}{\frac{3}{2}}-35\\ & =-8\end{array}$$