Answer: \(\mathrm{ff}(x)=x\); \(\mathrm f\) is self-inverse; range of \(\mathrm g\) is \(1\lt\mathrm g(x)\leqslant2\); domain of \(\mathrm h\) is \(x\ne-\frac13\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

The graph represents a function because each permitted vertical line \(x=k\) meets the graph in at most one point.

Now

\(\mathrm f(\mathrm f(x))=\frac{\frac{x}{x-1}}{\frac{x}{x-1}-1}.\)

The denominator is

\(\frac{x}{x-1}-1=\frac{x-(x-1)}{x-1}=\frac{1}{x-1}.\)

Therefore

\(\mathrm{ff}(x)=\frac{\frac{x}{x-1}}{\frac{1}{x-1}}=x.\)

So applying \(\mathrm f\) twice gives the identity function. Hence \(\mathrm f\) is its own inverse, so \(\mathrm f=\mathrm f^{-1}\).

On the diagram, this is shown by the graph being symmetrical about the line \(y=x\).

For \(\mathrm g(x)=\frac{x}{x-1}\), where \(x\geqslant2\), we have \(\mathrm g(2)=2\). As \(x\) increases, \(\frac{x}{x-1}=1+\frac{1}{x-1}\), so the values approach \(1\) from above.

Thus the range is \(1\lt\mathrm g(x)\leqslant2\).

For \(\mathrm h(x)=\frac{2x}{3x+1}\), the denominator must not be zero. Therefore \(3x+1\ne0\), so the domain is \(x\ne-\frac13\).

This gives the required answer: \(\mathrm{ff}(x)=x\); \(\mathrm f\) is self-inverse; range of \(\mathrm g\) is \(1\lt\mathrm g(x)\leqslant2\); domain of \(\mathrm h\) is \(x\ne-\frac13\).