Answer: \(5-(x+2)^2\); range \(\mathrm{f}(x)\leqslant5\); \(k=-2\); \(\mathrm{g}^{-1}(x)=-2+\sqrt{5-x}\), with domain \(x\leqslant5\) and range \(\mathrm{g}^{-1}(x)\geqslant-2\).

Work through the given conditions step by step, keeping exact values until the final answer is required.

Complete the square:

\(1-4x-x^2=-(x^2+4x-1)=5-(x+2)^2.\)

Since \((x+2)^2\geqslant0\), the greatest value of \(\mathrm{f}(x)\) is \(5\). Hence the range is \(\mathrm{f}(x)\leqslant5\).

For \(\mathrm{g}\) to have an inverse, its domain must be restricted to one side of the turning point. The least possible value is therefore \(k=-2\).

Now write \(y=5-(x+2)^2\), with \(x\geqslant-2\). Then

\((x+2)^2=5-y.\)

Since \(x+2\geqslant0\), \(x+2=\sqrt{5-y}\), so \(x=-2+\sqrt{5-y}\).

Interchanging \(x\) and \(y\),

\(\mathrm{g}^{-1}(x)=-2+\sqrt{5-x}.\)

The domain of \(\mathrm{g}^{-1}\) is \(x\leqslant5\), and the range is \(\mathrm{g}^{-1}(x)\geqslant-2\).

This gives the required answer: \(5-(x+2)^2\); range \(\mathrm{f}(x)\leqslant5\); \(k=-2\); \(\mathrm{g}^{-1}(x)=-2+\sqrt{5-x}\), with domain \(x\leqslant5\) and range \(\mathrm{g}^{-1}(x)\geqslant-2\).