And t 2 = x- 1, which means x = t 2+ 1.

So the edge of the original function is y = t 2+1-t.

=t^2-t+2

=(t- 1/2)^2+7/4(t≥0)

Therefore, when t= 1/2, y has a minimum value of 7/4, that is, y≥7/4.

So the value range of the function is [7/4, positive infinity].