Answer: \(\mathrm f(x)=15-2(x+1)^2\), range \(\mathrm f\leq15\); domain of \(\mathrm g^{-1}\): \(x\geq\sqrt2\), range: \(x\geq1\); \(\mathrm g^{-1}(x)=-1+\sqrt{x^2+2}\).

(a)(i) Complete the square:

\(\mathrm f(x)=13-4x-2x^2.\)

Factor out \(-2\) from the quadratic terms:

\(\mathrm f(x)=-2(x^2+2x)+13.\)

Since

\(x^2+2x=(x+1)^2-1,\)

we get

\(\mathrm f(x)=-2\left((x+1)^2-1\right)+13.\)

So

\(\mathrm f(x)=15-2(x+1)^2.\)

(a)(ii) Since \((x+1)^2\geq0\),

\(-2(x+1)^2\leq0.\)

Therefore

\(\mathrm f(x)\leq15.\)

(b)(i) The domain of \(\mathrm g\) is \(x\geq1\). Hence the range of \(\mathrm g^{-1}\) is

\(x\geq1.\)

Also

\(\mathrm g(1)=\sqrt{1+2-1}=\sqrt2,\)

and \(\mathrm g\) is increasing for \(x\geq1\). So the range of \(\mathrm g\), and hence the domain of \(\mathrm g^{-1}\), is

\(x\geq\sqrt2.\)

(b)(ii) Let

\(y=\sqrt{x^2+2x-1}.\)

Then

\(y^2=x^2+2x-1.\)

Rearrange as a quadratic in \(x\):

\(x^2+2x-(1+y^2)=0.\)

Using the quadratic formula,

\(x=\frac{-2\pm\sqrt{4+4(1+y^2)}}{2}.\)

So

\(x=-1\pm\sqrt{y^2+2}.\)

Since the range of \(\mathrm g^{-1}\) is \(x\geq1\), choose the positive branch:

\(x=-1+\sqrt{y^2+2}.\)

Replacing \(y\) by \(x\),

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