a problem:  plug in 0 or 1 and you get f(1)= 0 and f(1) = 1  so this is not a valid function over all real numbers.

Since √x appears, x>=0

Moreover, the function (x-√x +1) has a minimum at x = 1/4 seen by setting its derivative to zero:  1 - 1/(2*√x) = 0.  Plugging in x values >= 1/4 gives a unique value for the function.

So the domain of x is x >= 1/4.

Plugging in 1/4 above yields f(3/4) = 1/2

The domain of the function is 3/4.

Find x value such that (x-√x +1) = (21/16)

Could let y = √x; or just use quadratic formula and solve for √x

x - √x + 1 - 21/16 = 0

x - √x - 5/16 = 0

√x = (1 ± sqrt(9/4))/2

√x = {-1/4, 5/4} but reject the negative value since √x > 0

x = 25/16  plugging this into (x-√x +1) produces (21/16)

f(25/16-√(25/16) +1) = f(21/16) = √x = 5/4

Method Two:

let z = x - √x + 1

x - √x + 1 - z = 0

√x = (1 ± sqrt(1 - 4 + 4z))/2

   = (1 ± sqrt(4z-3))/2

The general solution for z > 3/4 is