11
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a2188eac-08f7-4e75-a76d-fe35b13a2e5f-4_1022_942_356_603}
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\caption{Fig. 11}
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Fig. 11 shows a sketch of the cubic curve \(y = \mathrm { f } ( x )\). The values of \(x\) where it crosses the \(x\)-axis are - 5 , - 2 and 2 , and it crosses the \(y\)-axis at \(( 0 , - 20 )\).
- Express \(\mathrm { f } ( x )\) in factorised form.
- Show that the equation of the curve may be written as \(y = x ^ { 3 } + 5 x ^ { 2 } - 4 x - 20\).
- Use calculus to show that, correct to 1 decimal place, the \(x\)-coordinate of the minimum point on the curve is 0.4 .
Find also the coordinates of the maximum point on the curve, giving your answers correct to 1 decimal place.
- State, correct to 1 decimal place, the coordinates of the maximum point on the curve \(y = \mathrm { f } ( 2 x )\).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a2188eac-08f7-4e75-a76d-fe35b13a2e5f-5_689_1006_269_568}
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\caption{Fig. 12}
\end{figure}
A water trough is a prism 2.5 m long. Fig. 12 shows the cross-section of the trough, with the depths in metres at 0.1 m intervals across the trough. The trough is full of water. - Use the trapezium rule with 5 strips to calculate an estimate of the area of cross-section of the trough.
Hence estimate the volume of water in the trough.
- A computer program models the curve of the base of the trough, with axes as shown and units in metres, using the equation \(y = 8 x ^ { 3 } - 3 x ^ { 2 } - 0.5 x - 0.15\), for \(0 \leqslant x \leqslant 0.5\).
Calculate \(\int _ { 0 } ^ { 0.5 } \left( 8 x ^ { 3 } - 3 x ^ { 2 } - 0.5 x - 0.15 \right) \mathrm { d } x\) and state what this represents.
Hence find the volume of water in the trough as given by this model.