Answer: \(2\left(x-\frac12\right)^2+\frac52\), \(p=\frac12\), range \(\mathrm{f}(x)\geqslant\frac52\), and \(\mathrm{f}^{-1}(x)=\frac12-\sqrt{\frac{x-\frac52}{2}}\).

Work through the given conditions step by step, keeping exact values until the final answer is required.

Complete the square:

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

Since \(x^2-x=\left(x-\frac12\right)^2-\frac14\),

\(2x^2-2x+3=2\left(x-\frac12\right)^2-\frac12+3.\)

So

\(2x^2-2x+3=2\left(x-\frac12\right)^2+\frac52.\)

The vertex is at \(x=\frac12\). For the restricted domain \(x\leqslant p\), the greatest value of \(p\) for which \(\mathrm{f}\) is one-one is

\(p=\frac12.\)

The minimum value is \(\frac52\), so the range is

\(\mathrm{f}(x)\geqslant\frac52.\)

Let \(y=2\left(x-\frac12\right)^2+\frac52\). Then

\(\frac{y-\frac52}{2}=\left(x-\frac12\right)^2.\)

Since the domain is \(x\leqslant\frac12\), take the negative square root:

\(x-\frac12=-\sqrt{\frac{y-\frac52}{2}}.\)

Hence

\(\mathrm{f}^{-1}(x)=\frac12-\sqrt{\frac{x-\frac52}{2}}.\)

This gives the required answer: \(2\left(x-\frac12\right)^2+\frac52\), \(p=\frac12\), range \(\mathrm{f}(x)\geqslant\frac52\), and \(\mathrm{f}^{-1}(x)=\frac12-\sqrt{\frac{x-\frac52}{2}}\).