Where a function meets its inverse
Where a function meets its inverse
10 marks
TGM Original
The function \(\mathrm{f}\) is defined by \(\mathrm{f}(x)=x^2-2x+2\), for \(x\ge1\).
- Find an expression for \(\mathrm{f}^{-1}(x)\), and state the domain and range of \(\mathrm{f}^{-1}\).[2]
- Given that \(\mathrm{f}\) is increasing, explain why any point where the graphs of \(y=\mathrm{f}(x)\) and \(y=\mathrm{f}^{-1}(x)\) meet must lie on the line \(y=x\).[1]
- Hence find the coordinates of the points where these two graphs meet.[2]
- Sketch \(y=\mathrm{f}(x)\), \(y=\mathrm{f}^{-1}(x)\) and \(y=x\) on the same axes, using equal scales. Label each graph and the points found in (c).[2]
The function \(\mathrm{g}\) is defined by \(\mathrm{g}(x)=(1-x)^2\), for \(0\le x\le1\). It is decreasing.
- Show that the points \((0,1)\) and \((1,0)\) lie on both \(y=\mathrm{g}(x)\) and \(y=\mathrm{g}^{-1}(x)\). Solve \(\mathrm{g}(x)=x\), and explain why solving \(\mathrm{g}(x)=x\) does not find every point where these two graphs meet.[3]
Solution:
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Answer:(a) \(\mathrm{f}^{-1}(x)=1+\sqrt{x-1}\), domain \(x\ge1\), range \(\mathrm{f}^{-1}(x)\ge1\) (b) \(\mathrm{f}(a)=b\) and \(\mathrm{f}(b)=a\) with \(a\ne b\) contradicts \(\mathrm{f}\) increasing (c) \((1,1)\), \((2,2)\) (d) Sketch: reflections in \(y=x\) meeting at \((1,1)\) and \((2,2)\) (e) \(x=\dfrac{3-\sqrt5}{2}\); the graphs also meet at \((0,1)\) and \((1,0)\), off \(y=x\)
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