Answer: (a)(i) domain \(0\leq x\leq\frac94\), range \(0\leq f^{-1}(x)\leq3\); (a)(ii) \(x=0\) or \(x=\frac85\); (b)(i) \(g^{-1}(x)=\sqrt[3]{\frac{x^3-3}{8}}\); (b)(ii) \(k=0\); (b)(iii) \(gh(x)=\sqrt[3]{8e^{12x}+3}\).

For an inverse function, interchange \(x\) and \(y\), then rearrange to make \(y\) the subject. The domain of the original function becomes the range of the inverse.

(a)(i) The domain of \(f^{-1}\) is the range of \(f\). Since

\(f(0)=0\)

and

\(f(3)=\frac{9}{\sqrt{16}}=\frac94,\)

the domain of \(f^{-1}\) is

\(0\leq x\leq\frac94.\)

The range of \(f^{-1}\) is the domain of \(f\), so

\(0\leq f^{-1}(x)\leq3.\)

(a)(ii) Solve

\(\frac{3x}{\sqrt{5x+1}}=x.\)

One solution is \(x=0\). If \(x\ne0\), divide by \(x\):

\(\frac{3}{\sqrt{5x+1}}=1.\)

Hence

\(\sqrt{5x+1}=3.\)

Squaring gives

\(5x+1=9,\)

so

\(x=\frac85.\)

Thus

\(x=0\quad\text{or}\quad x=\frac85.\)

(a)(iii) The graph of \(y=f^{-1}(x)\) is the reflection of the graph of \(y=f(x)\) in the line \(y=x\). It passes through \((0,0)\) and \(\left(\frac94,3\right)\), and it intersects the graph of \(y=f(x)\) at \((0,0)\) and \(\left(\frac85,\frac85\right)\).

(b)(i) Let

\(y=\sqrt[3]{8x^3+3}.\)

Then

\(y^3=8x^3+3.\)

So

\(x^3=\frac{y^3-3}{8}.\)

Therefore

\(x=\sqrt[3]{\frac{y^3-3}{8}}.\)

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

\(g^{-1}(x)=\sqrt[3]{\frac{x^3-3}{8}}.\)

(b)(ii) For \(gh(x)\) to be formed, the output of \(h\) must be in the domain of \(g\). Since the domain of \(g\) is \(x\geq1\), we need

\(e^{4x}\geq1.\)

This is true when \(x\geq0\), so the least possible value is

\(k=0.\)

(b)(iii) Now

\(gh(x)=g(h(x))=g(e^{4x}).\)

Therefore

\(gh(x)=\sqrt[3]{8(e^{4x})^3+3} =\sqrt[3]{8e^{12x}+3}.\)