Answer:

(b) \(a=1\), range \(\mathrm h(x)\ge3\), \(\mathrm h^{-1}(x)=1+\sqrt{x-3}\), domain \(x\ge3\).

Answer: \(a=1\), \(\mathrm h(x)\ge3\), \(\mathrm h^{-1}(x)=1+\sqrt{x-3}\), with domain \(x\ge3\).

For \(\mathrm f(x)=2x^2\),

\(\mathrm f'(x)=4x.\)

Also,

\(\mathrm{fg}(x)=\mathrm f(\mathrm g(x))=2(2x+1)^2=8x^2+8x+2.\)

Next,

\(\mathrm g^2(x)=\mathrm g(\mathrm g(x))=\mathrm g(2x+1)=2(2x+1)+1=4x+3.\)

Finally, from \(y=2x+1\),

\(x=\frac{y-1}{2}.\)

So

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

For part (b),

\(\mathrm h(x)=(x-1)^2+3.\)

The turning point is at \(x=1\). For \(\mathrm h^{-1}\) to exist with \(a\) as small as possible, use the domain

\(x\ge1.\)

So \(a=1\). The least value of \(\mathrm h(x)\) is \(3\), hence the range is

\(\mathrm h(x)\ge3.\)

To find the inverse, write

\(y=(x-1)^2+3.\)

Then

\(x-1=\sqrt{y-3}\)

because \(x\ge1\). Thus

\(x=1+\sqrt{y-3}.\)

Therefore

\(\mathrm h^{-1}(x)=1+\sqrt{x-3}.\)

The domain of \(\mathrm h^{-1}\) is the range of \(\mathrm h\), so

\(x\ge3.\)