This question is not difficult. You just need to write down the coordinates of two points in MN.

Let M (x, y) and then transform f to get N(y, x).

| Mn | 2 = (x-y) 2+(y-x) 2 = 2 (y-x) 2, with y 2 = x-1eliminate x, and solve it as a unary function.

|mn|^2=2(y-y^2- 1)^2=2[(y- 1/2)^2+3/4]^2，

When y= 1/2, the minimum value of | Mn | 2 =9/8.

Then the minimum value of |MN| is 3√2/4.

If two curves are hard to imagine, you can also turn them into a curve and a straight line to solve.

A and A 1 are symmetrical about the straight line y=x, and curves c and C 1 are also symmetrical about the straight line y = X.

As long as we find the shortest distance from c to the straight line y=x, multiplied by 2, it is the minimum value of MN.

I'm so sleepy. Anyway, the formulas behind the two methods are similar, so I won't write them.