Composites fg and gf with their domains
Composites fg and gf with their domains
8 marks
TGM Original
The functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by \(\mathrm{f}(x)=\dfrac{1}{x-2}\), for \(x>2\), and \(\mathrm{g}(x)=x^2+1\), for \(x\in\mathbb{R}\).
- Find expressions for \(\mathrm{fg}(x)\) and \(\mathrm{gf}(x)\).[2]
- Evaluate \(\mathrm{fg}(3)\) and \(\mathrm{gf}(3)\), and state what this shows.[1]
- State the domain of \(\mathrm{gf}\), explaining why it lies within the domain of \(\mathrm{f}\).[1]
- Find the range of \(\mathrm{gf}\), and explain why it lies within the range of \(\mathrm{g}\).[2]
- Explain why \(\mathrm{fg}(x)\) cannot be formed for every real \(x\). The domain of \(\mathrm{g}\) is restricted to \(x>k\) so that \(\mathrm{fg}\) exists. Find the least possible value of \(k\).[2]
Solution:
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Answer:(a) \(\mathrm{fg}(x)=\dfrac{1}{x^2-1}\), \(\mathrm{gf}(x)=\dfrac{1}{(x-2)^2}+1\) (b) \(\dfrac18\ne2\), so \(\mathrm{fg}\ne\mathrm{gf}\) (c) \(x>2\) (d) \(\mathrm{gf}(x)>1\) (e) \(k=1\)
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