Answer: (b) \(\mathrm{gf}(x)=\dfrac{(4x-1)^2}{9x^4}\); (c) \(\mathrm f^{-1}(x)=\dfrac{2x-\sqrt{x(4x-3)}}{3}\).

We start with the main method. Check whether the function is one-to-one on the stated domain before finding an inverse.

The function \(\mathrm g\) has domain \(x\lt 0\), and

\(\mathrm g(x)=\frac{1}{x^2}\).

For \(x\lt 0\), \(\mathrm g(x)\gt 0\). However, \(\mathrm f\) is defined only for inputs less than \(0\).

Therefore \(\mathrm{fg}\) does not exist, because the range of \(\mathrm g\) is not inside the domain of \(\mathrm f\).

For part (b),

\(\mathrm{gf}(x)=\mathrm g(\mathrm f(x))=\frac{1}{(\mathrm f(x))^2}.\)

Since

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

we get

\(\mathrm{gf}(x)=\frac{1}{\left(\frac{3x^2}{4x-1}\right)^2} =\frac{(4x-1)^2}{9x^4}.\)

For part (c), let

\(y=\frac{3x^2}{4x-1}.\)

Then

\(y(4x-1)=3x^2.\)

So

\(3x^2-4xy+y=0.\)

Now treat this as a quadratic equation in \(x\):

\(x=\frac{4y\pm\sqrt{16y^2-12y}}{6}.\)

Hence

\(x=\frac{2y\pm\sqrt{y(4y-3)}}{3}.\)

Since the original domain of \(\mathrm f\) is \(x\lt 0\), we choose the negative square-root branch.

Therefore

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

This completes the solution and gives the required result.