Answer: range \(\mathrm f(x)\gt2\); \(\mathrm f^{-1}(x)=-\frac13\ln(x-2)\); asymptotes \(y=2\) and \(x=2\); \(\mathrm{gf}(x)=12\) gives \(x=-\frac13\ln2\).

We start with the main method. Use the measurements and relationships shown in the diagram, then translate them into algebraic or trigonometric equations. Use the fact that the graph of an inverse function is the reflection in the line \(y=x\).

(a) Since \(e^{-3x}\gt0\) for all real \(x\),

\(\mathrm f(x)=2+e^{-3x}\gt2.\)

So the range is

\(\mathrm f(x)\gt2.\)

(b) Let

\(y=2+e^{-3x}.\)

Then

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

Taking natural logarithms gives

\(\ln(y-2)=-3x.\)

So

\(x=-\frac13\ln(y-2).\)

Therefore

\(\mathrm f^{-1}(x)=-\frac13\ln(x-2),\quad x\gt2.\)

(c) The graph of \(y=\mathrm f(x)\) has \(y\)-intercept

\(2+e^0=3,\)

so it passes through \((0,3)\). Its horizontal asymptote is

\(y=2.\)

The graph of \(y=\mathrm f^{-1}(x)\) is the reflection of \(y=\mathrm f(x)\) in the line \(y=x\). It has \(x\)-intercept \((3,0)\) and vertical asymptote

\(x=2.\)

(d) The equation \(\mathrm{gf}(x)=12\) means

\((\mathrm f(x))^{3/2}+4=12.\)

So

\((2+e^{-3x})^{3/2}=8.\)

Raise both sides to the power \(\frac23\):

\(2+e^{-3x}=4.\)

Thus

\(e^{-3x}=2.\)

Taking natural logarithms,

\(-3x=\ln2,\)

so

\(x=-\frac13\ln2.\)

This completes the solution and gives the required result.