#### Answer

$\frac{7}{8}$

#### Work Step by Step

$D(x) = \sqrt {((x^2 + 1) - (x + x^2))^2}$
$D(x) = x^2 + 1 - x + x^2$
$D(x) = 2x^2 - x +1$
$D'(x) = 4x - 1$
$0 = 4x -1$
$4x = 1$
$x = \frac{1}{4}$
2($\frac{1}{4})^2 - \frac{1}{4}+1 = \frac{7}{8}$
There is only one value, so this is the minimum.