## Calculus: Early Transcendentals 8th Edition

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