Answer: The graph of \(y=\mathrm{f}^{-1}(x)\) is the reflection of \(y=\mathrm{f}(x)\) in the line \(y=x\). Also \(\mathrm{g}^{-1}(x)=(\ln x)^2+2\), the range of \(\mathrm{g}^{-1}\) is \([2,\infty)\), and \(\mathrm{gh}(x)=\mathrm{e}^{1/x}\).

(a) To sketch \(y=\mathrm{f}^{-1}(x)\), interchange the coordinates of points on the graph of \(y=\mathrm{f}(x)\). For example, a point \((a,b)\) on \(y=\mathrm{f}(x)\) becomes \((b,a)\) on \(y=\mathrm{f}^{-1}(x)\).

This is exactly the same as reflecting the graph of \(y=\mathrm{f}(x)\) in the line

\(y=x\).

Therefore the relationship is:

\(y=\mathrm{f}^{-1}(x)\) is the reflection of \(y=\mathrm{f}(x)\) in \(y=x\).

(b)(i) Let

\(y=\mathrm{g}(x)=\mathrm{e}^{\sqrt{x-2}}\), where \(x\geqslant2\).

Take natural logarithms:

\(\ln y=\sqrt{x-2}\).

Now square both sides:

\((\ln y)^2=x-2\).

So

\(x=(\ln y)^2+2\).

Interchange \(x\) and \(y\) to obtain the inverse function:

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

Since \(\mathrm{g}(x)\geqslant1\), the domain of \(\mathrm{g}^{-1}\) is \(x\geqslant1\).

(b)(ii) The range of \(\mathrm{g}^{-1}\) is the original domain of \(\mathrm{g}\). Since the domain of \(\mathrm{g}\) is \(x\geqslant2\),

\(\text{range of }\mathrm{g}^{-1}=[2,\infty)\).

(b)(iii) The notation \(\mathrm{gh}(x)\) means

\(\mathrm{g}(\mathrm{h}(x))\).

Now

\(\mathrm{h}(x)=\frac1{x^2}+2\), where \(x\gt0\). Hence

\(\mathrm{gh}(x)=\mathrm{g}\left(\frac1{x^2}+2\right)\).

Using \(\mathrm{g}(x)=\mathrm{e}^{\sqrt{x-2}}\),

\(\mathrm{gh}(x)=\mathrm{e}^{\sqrt{\left(\frac1{x^2}+2\right)-2}}\).

This simplifies to

\(\mathrm{gh}(x)=\mathrm{e}^{\sqrt{1/x^2}}\).

Since \(x\gt0\), \(\sqrt{1/x^2}=1/x\). Therefore

\(\mathrm{gh}(x)=\mathrm{e}^{1/x}\).