Inequalities

Example 0..1 Solve the inequality

$\displaystyle \frac{1}{2}(1 + x) \leq 6$

SOLUTION To solve this inequality, we first multiply both sides of inequality by 2.

$\displaystyle 1 + x \leq 12$

Then subtract 1 from both sides of inequality to obtain

$\displaystyle x \leq 11$

Then the solution is $(-\infty, 11]$ ?D $\ \blacksquare$

Example 0..2 Solve the inequality?D

$\displaystyle \frac{1}{5}(x^{2} - 4x + 3) < 0$

SOLUTION To solve this inequality, we first multiply both sides of inequality by 5.

$\displaystyle x^{2} - 4x + 3 < 0$

Then factor the left-hand side to obtain

$\displaystyle (x-1)(x-3) < 0$

Note that becomes 0 at 1 and 3?DThen we make circle on the number line at 1 and 3. Then we have 3 separated parts?D

$\displaystyle (-\infty, 1),\ (1,3), \ (3, \infty)$

In the given interval?Cthe sign of the product does not change. Thus we have

$(-\infty, 1)$	${\rm sgn}[(x-1)(x-3)] = (-)(-) = +$
	${\rm sgn}[(x-1)(x-3)] = (+)(-) = -$
$(3, \infty)$	${\rm sgn}[(x-1)(x-3)] = (+)(+) = +$

From this we obtain ?D $\ \blacksquare$

Example 0..3 Solve the inequality?D

$\displaystyle \frac{x+2}{1 - x} \geq 1$

SOLUTION To solve this inequality, we add to both sides of the inequality to get rid of 1.

$\displaystyle \frac{x+2}{1 - x} - 1 \geq 0$

Simplifying

$\displaystyle \frac{x+2 - (1 - x)}{1 - x}$	$\displaystyle \geq$	0
$\displaystyle \frac{2x + 1}{1-x} \geq 0$

Now we would like to get rid of the denominator . But if we multiply to get rid of the denominator, then we have to be careful about the sign of . Thus instead?Cwe multiply $(1-x)^{2}$ . Then we have

$\displaystyle (2x+1)(1-x) \geq 0$

Now as above, the product becomes 0 at $-\frac{1}{2}$ and 1?DThen we make circle on the number line at $-\frac{1}{2}$ and 1?D Next note that the equality only holds at the point where the numerator is 0. From this we fill the circle at $-\frac{1}{2}$ ?DFrom this the number line can be separated by 3 parts?D

$\displaystyle (-\infty, -\frac{1}{2}],\ [-\frac{1}{2},1), \ (1, \infty)$

$(-\infty, -\frac{1}{2}]$	${\rm sgn}[(2x+1)(1-x)] = (-)(+) = -$
$[-\frac{1}{2},1)$	${\rm sgn}[(2x+1)(1-x)] = (+)(+) = +$
$(1, \infty)$	${\rm sgn}[(2x+1)(1-x)] = (+)(-) = -$

Therefore, the solution is $[-\frac{1}{2},1)$ ?D $\ \blacksquare$