Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - Chapter Review Exercises - Page 37: 32


$$ x^{2}+3 x+3\geq 0,\ \ \text{for all } x$$

Work Step by Step

We can rewrite the expression as: \begin{align*} x^{2}+3 x+3&=\left(x+\frac{3}{2}\right)^2-\frac{9}{4} +3\\ &= \left(x-\frac{3}{2}\right)^{2}+\frac{3}{4} \end{align*} We see that $\left(x-\frac{3}{2}\right)^{2} \geq 0$ for all $x$, so $$ x^{2}+3 x+3\geq 0,\ \ \text{for all } x$$
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