Answer: (i) \(f(x)=2\left(x-\dfrac32\right)^2+\dfrac12\). (ii) The graphs are reflections in the line \(y=x\). (iii) \(f^{-1}(x)=\dfrac32-\sqrt{\dfrac{x-\frac12}{2}}\), with domain \(x\geqslant \dfrac12\).

(i) Complete the square:

\(f(x)=2x^2-6x+5 =2(x^2-3x)+5.\)

Now

\(x^2-3x=\left(x-\frac32\right)^2-\frac94,\)

so

\(f(x)=2\left[\left(x-\frac32\right)^2-\frac94\right]+5 =2\left(x-\frac32\right)^2-\frac92+5 =2\left(x-\frac32\right)^2+\frac12.\)

(ii) The graph of \(y=f(x)\) is the left-hand branch of the parabola with vertex \(\left(\frac32,\frac12\right)\), because the domain is \(x\leqslant\frac32\).

The graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\). In particular, the vertex \(\left(\frac32,\frac12\right)\) reflects to \(\left(\frac12,\frac32\right)\).

(iii) Start from

\(y=2\left(x-\frac32\right)^2+\frac12.\)

Swap \(x\) and \(y\):

\(x=2\left(y-\frac32\right)^2+\frac12.\)

Then

\(\frac{x-\frac12}{2}=\left(y-\frac32\right)^2.\)

Taking square roots gives

\(y-\frac32=\pm\sqrt{\frac{x-\frac12}{2}}.\)

Because the original domain is \(x\leqslant\frac32\), the inverse must use the branch with \(y\leqslant\frac32\). So we take the negative root:

\(f^{-1}(x)=\frac32-\sqrt{\frac{x-\frac12}{2}}.\)

The domain of \(f^{-1}\) is the range of \(f\), which is

\(x\geqslant\frac12.\)