Answer: \(\mathrm{f}^2(x)=4x-3\), intercepts \((\frac34,0)\) and \((0,3)\), \(a=\frac12\), \(b=\frac{15}{2}\).

(a) The range of \(\mathrm{f}\) is all real numbers, and this is the same as the domain of \(\mathrm{f}\). Therefore \(\mathrm{f}^2=\mathrm{f}\mathrm{f}\) can be formed.

(b) First find \(\mathrm{f}^2(x)\):

\(\mathrm{f}^2(x)=\mathrm{f}(\mathrm{f}(x))=2(2x-1)-1=4x-3.\)

So

\(y=\left|\mathrm{f}^2(x)\right|=|4x-3|.\)

This is a V-shaped graph with vertex where \(4x-3=0\), so the \(x\)-intercept is

\(\left(\frac34,0\right).\)

The \(y\)-intercept is found by putting \(x=0\):

\(y=|-3|=3,\)

so the \(y\)-intercept is \((0,3)\).

(c) The inequality holds exactly for \(-1\leqslant x\leqslant3\), so the line \(y=ax+b\) passes through the corresponding end points on \(y=|4x-3|\).

When \(x=-1\),

\(|4(-1)-3|=7.\)

When \(x=3\),

\(|4(3)-3|=9.\)

So the line passes through \((-1,7)\) and \((3,9)\). Its gradient is

\(a=\frac{9-7}{3-(-1)}=\frac12.\)

Using \((-1,7)\),

\(7=-\frac12+b,\)

so

\(b=\frac{15}{2}.\)