Answer: \(\mathrm{fg}(x)=\frac{2}{3x}-\frac{x}{3}\); range of \(\mathrm f^{-1}\) is \(x\gt 0\); \(\mathrm f^{-1}(x)=\frac{3x+\sqrt{9x^2+8}}{4}\).

(a) Since \(\mathrm g(x)=\frac1x\),

Substitute \(\frac1x\) into \(\mathrm f\):

So

Therefore

\(\mathrm{fg}(x)=\frac{2}{3x}-\frac{x}{3}.\)

(b)(i) The domain of \(\mathrm f\) is \(x\gt 0\). Therefore the range of \(\mathrm f^{-1}\) is

\(x\gt 0.\)

(b)(ii) Let

\(y=\frac{2x^2-1}{3x}.\)

Then

\(3xy=2x^2-1.\)

Rearrange as a quadratic in \(x\):

\(2x^2-3yx-1=0.\)

Using the quadratic formula,

\(x=\frac{3y\pm\sqrt{9y^2+8}}{4}.\)

Since \(x\gt 0\), take the positive square-root branch:

\(x=\frac{3y+\sqrt{9y^2+8}}{4}.\)

Replacing \(y\) by \(x\),

\(\mathrm f^{-1}(x)=\frac{3x+\sqrt{9x^2+8}}{4}.\)