C3 January 2008 Q8
8. The functions f and g are defined by\[\mathrm{f} : x \mapsto 1 - 2x^3, \ x \in \mathbb{R}\]\[\mathrm{g} : x \mapsto \frac{3}{x} - 4, \ x > 0, \ x \in \mathbb{R}\]
(a) Find the inverse function \(\mathrm{f}^{-1}\). (2)
(b) Show that the composite function gf is\[\mathrm{gf} : x \mapsto \frac{8x^3 - 1}{1 - 2x^3}.\] (4)
(c) Solve \(\mathrm{gf}(x) = 0\). (2)
(d) Use calculus to find the coordinates of the stationary point on the graph of \(y = \mathrm{gf}(x)\). (5)
| Scheme | Marks |
|---|---|
| \(x = 1 - 2y^3 \Rightarrow y = \left(\dfrac{1 - x}{2}\right)^{\frac{1}{3}}\) or \(\sqrt[3]{\dfrac{1 - x}{2}}\) | M1 A1 |
| \(\mathrm{f}^{-1} : x \mapsto \left(\dfrac{1 - x}{2}\right)^{\frac{1}{3}}\) Ignore domain | |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathrm{gf}(x) = \dfrac{3}{1 - 2x^3} - 4\) | M1 A1 |
| \(= \dfrac{3 - 4(1 - 2x^3)}{1 - 2x^3}\) | M1 |
| \(= \dfrac{8x^3 - 1}{1 - 2x^3}\ \ \ast\) cso | A1 |
| \(\mathrm{gf} : x \mapsto \dfrac{8x^3 - 1}{1 - 2x^3}\) Ignore domain | |
| (4) |
| Scheme | Marks |
|---|---|
| \(8x^3 - 1 = 0\) Attempting solution of numerator \(= 0\) | M1 |
| \(x = \dfrac{1}{2}\) Correct answer and no additional answers | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{(1 - 2x^3)\times 24x^2 + (8x^3 - 1)\times 6x^2}{(1 - 2x^3)^2}\) | M1 A1 |
| \(= \dfrac{18x^2}{(1 - 2x^3)^2}\) | A1 |
| Solving their numerator \(= 0\) and substituting to find \(y\). | M1 |
| \(x = 0,\ y = -1\) | A1 |
| (5) | |
| (13 marks) |