Chapter 8 - Section 8.4 - Nonlinear Inequalities in One Variable - Exercise Set - Page 511: 66

$(0, ∞)$

$2x+1/x \ge 0$ $x(2x+1/x) \ge 0*x$ $2x^2+1 \ge 0$ $2x^2+1-1 \ge 0-1$ $2x^2 \ge -1$ $2x^2/2 \ge -1/2$ $x^2 \ge -1/2$ $\sqrt {x^2} = \sqrt {-1/2}$ $x = \sqrt {-1} /\sqrt {2}$ $x= i*\sqrt 2 /\sqrt 2 *\sqrt 2$ $x=±i\sqrt2 /2$ Disregard $i$ since we are working in the real numbers. We then have $x= ±\sqrt 2/2$ The denominator of the original equation is zero when $x=0$ We have four ranges, and we need to test one value from each range to see if the range is part of the solution set. These ranges are $(-∞, -\sqrt2/2)$, $(-\sqrt 2/2, 0)$, $(0, \sqrt 2/2)$, and $(\sqrt 2/2,∞)$. Since we don't have a "less than or equal to" sign or a "greater than or equal to" sign, we exclude the endpoints and use parentheses and not brackets. Let $x=-2$, $x=-.5$, $x=.5$, $x=2$ $x=-2$ $2x+1/x \ge 0$ $2*-2+1/-2 \ge 0$ $-4-1/2\ge 0$ $-9/2 \ge 0$ (false) $x=-.5$ $2x+1/x \ge 0$ $2*-.5+1/-.5 \ge 0$ $-1-2 \ge 0$ $-3 \ge 0$ (false) $x=.5$ $2x+1/x \ge 0$ $2*.5+1/.5 \ge 0$ $1 + 2 \ge 0$ $3 \ge 0$ (true) $x=2$ $2x+1/x \ge 0$ $2*2+1/2 \ge 0$ $4+1/2 \ge0$ $9/2 \ge 0$ (true)