Answer: \(g^{-1}(x)=4-\sqrt{x-1}\), domain \(x\geq1\), range \(g^{-1}(x)\leq4\); \(fg(x)=\ln(2x^2-16x+35)\).
(a) The graph of
\(f(x)=\ln(2x+1),\qquad x\geq0,\)
starts at
\(f(0)=\ln1=0,\)
and then increases, with gradually decreasing gradient.
The graph of \(y=f^{-1}(x)\) is the reflection of the graph of \(y=f(x)\) in the line \(y=x\). It also passes through \((0,0)\), but has the corresponding inverse shape.
(b)(i) Let
\(y=(x-4)^2+1.\)
Then
\(y-1=(x-4)^2.\)
Since \(x\leq4\), we must take the negative square root:
\(x-4=-\sqrt{y-1}.\)
So
\(x=4-\sqrt{y-1}.\)
Therefore
\(g^{-1}(x)=4-\sqrt{x-1}.\)
The range of \(g\) is \(g(x)\geq1\), so the domain of \(g^{-1}\) is
\(x\geq1.\)
The domain of \(g\) is \(x\leq4\), so the range of \(g^{-1}\) is
\(g^{-1}(x)\leq4.\)
(b)(ii) We have
\(fg(x)=f(g(x)).\)
Since \(f(x)=\ln(2x+1)\),
\(f(g(x))=\ln(2g(x)+1).\)
Substitute \(g(x)=(x-4)^2+1\):
\(fg(x)=\ln\left(2\left[(x-4)^2+1\right]+1\right).\)
Simplify the expression inside the logarithm:
\(2\left[(x-4)^2+1\right]+1 =2(x^2-8x+16)+3 =2x^2-16x+35.\)
Therefore
\(fg(x)=\ln(2x^2-16x+35).\)
(b)(iii) The function \(gf\) would mean \(g(f(x))\). For this to exist for the whole domain of \(f\), every output of \(f\) must lie in the domain of \(g\).
But \(f(x)=\ln(2x+1)\) is unbounded above, while the domain of \(g\) is \(x\leq4\). Some values in the range of \(f\) are greater than \(4\), so they are outside the domain of \(g\).
Therefore \(gf\) does not exist.
