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1+\frac{x}{2} -\frac{x^2}{8} \le \sqrt{1+x} \le 1+\frac{x}{2}

Source: Germany 1999 p2

February 23, 2020

inequalitiesalgebra

Problem Statement

Determine all real numbers

x

for which

1+\frac{x}{2} -\frac{x^2}{8} \le \sqrt{1+x} \le 1+\frac{x}{2}