Answer: range of \(\mathrm f\): \(\mathrm f(x)\geq -4\); range of \(\mathrm g\): \(\mathrm g(x)\gt1\); \(x=\ln\sqrt2\).

(a) Complete the square:

\(x^2+4x=(x+2)^2-4.\)

Since \((x+2)^2\geq0\), the least value of \(\mathrm f(x)\) is \(-4\). Hence the range of \(\mathrm f\) is

\(\mathrm f(x)\geq -4.\)

(b) Since \(\mathrm e^{2x}\gt0\) for all real \(x\),

\(\mathrm g(x)=1+\mathrm e^{2x}\gt1.\)

So the range of \(\mathrm g\) is

\(\mathrm g(x)\gt1.\)

(c) The equation \(\mathrm{fg}(x)=21\) means

\(\mathrm f(\mathrm g(x))=21.\)

Substitute \(\mathrm g(x)=1+\mathrm e^{2x}\) into \(\mathrm f(u)=u^2+4u\):

\((1+\mathrm e^{2x})^2+4(1+\mathrm e^{2x})=21.\)

Expand:

So

\(\mathrm e^{4x}+6\mathrm e^{2x}-16=0.\)

Let \(u=\mathrm e^{2x}\). Then

\(u^2+6u-16=0.\)

Factorise:

\((u+8)(u-2)=0.\)

Since \(u=\mathrm e^{2x}\gt0\), take \(u=2\). Hence

\(\mathrm e^{2x}=2.\)

Taking logarithms,

\(2x=\ln2,\)

so

\(x=\frac12\ln2=\ln\sqrt2.\)