It is given that
\({\mathrm{f}:x\mapsto\begin{cases} (x-2)^2, & 0 < x \leq 3, \\ 3x-8, & 3 < x \leq 4, \end{cases}}\)
Sketch the graph of \(y=\mathrm{f}\left( x \right)\), labelling the coordinates of any turning points and endpoints. Explain why \({{\mathrm{f}}^{-1}}\) does not exist.[3]
If the domain of \(\mathrm{f}\) is restricted to \(\left( 0,k \right]\), state the largest value of \(k\) such that \({{\mathrm{f}}^{-1}}\) exists. Hence, for this value of \(k\), find \({{\mathrm{f}}^{-1}}\left( x \right)\) and state the domain of \({{\mathrm{f}}^{-1}}\).[3]
The function \(\mathrm{g}\) is such that
\(g:x\mapsto {{\mathrm{e}}^{x}}+3\), \(x\le 0\).
Find the function \(\mathrm{fg}\), giving your answer in similar form.[3]
It is given further that \(\mathrm{f}\left( x \right)=\mathrm{f}\left( x+4 \right)\).
Evaluate \(\mathrm{f}\left( 25 \right)\) and \(\mathrm{f}\left( -8 \right)\).[2]
Sketch the graph of \(y=\mathrm{f}\left( \frac{1}{2}x-1 \right)\) for \(-8 < x\le 10\).[2]