Answer: (i) \(g'(x)=-\dfrac{6}{(2x+1)^2}\), so it is always negative. (ii) \(g(x)\gt 0\). (iii) \(hg(x)=\dfrac{3k}{2x+1}+3\). (iv) \(k=\dfrac23\). (v) Domain: \(x\gt -\dfrac12\).

(i) Write \(g(x)\) as

\(g(x)=3(2x+1)^{-1}.\)

Differentiate:

\(g'(x)=3(-1)(2x+1)^{-2}\cdot2 =-6(2x+1)^{-2} =-\frac{6}{(2x+1)^2}.\)

Since \((2x+1)^2\gt 0\) for every \(x\) in the domain, the denominator is always positive, and the numerator is negative. Therefore \(g'(x)\) is always negative.

(ii) Because \(x\gt -\frac12\), we have \(2x+1\gt 0\), so

\(g(x)=\frac{3}{2x+1}\gt 0.\)

Hence the range of \(g\) is

\(g(x)\gt 0.\)

(iii) Since \(h(x)=kx+3\),

\(hg(x)=h(g(x))=k\left(\frac{3}{2x+1}\right)+3=\frac{3k}{2x+1}+3.\)

(iv) Substitute \(x=0\):

\(hg(0)=\frac{3k}{2(0)+1}+3=3k+3.\)

Given that \(hg(0)=5\),

\(3k+3=5.\)

So

\(3k=2 \quad\Rightarrow\quad k=\frac23.\)

(v) The composite \(hg\) is defined wherever \(g\) is defined, so its domain is

\(x\gt -\frac12.\)