College Algebra - Rational Inequalities

The boundary points of a rational inequality are the zeroes of the numerator and the denominator.

First, set the numerator equal to 0 and solve for :

Now set the denominator equal to 0 and solve for :

$x^{2}-8x + 7=0$

Factor the trinomial using the reverse-FOIL method. Look for two integers whose sum is and whose product is 7; these are , and , so the equation can be rewritten as

Set each factor to zero and solve for :

Therefore, the boundary points are $\left { 0,1,7\right }$ , which divide the real numbers into four intervals. Choose any value from each interval as a test point, setting to that value and determining whether the inequality is true.

The four intervals are listed below, along with their arbitrary test points.

$(-\infty, 0)$ : Set

$\frac{4x}{x^{2}-8x + 7} \le 0$

$\frac{4 (-1)}{(-1)^{2}-8(-1) + 7} \le 0$

$\frac{-4}{16} \le 0$

$-\frac{1}{4} \le 0$

True; include $(-\infty, 0)$ .

: Set $x= \frac{1}{2}$

$\frac{4x}{x^{2}-8x + 7} \le 0$

$\frac{4\cdot \frac{1}{2}}{ \left (\frac{1}{2} \right )^{2}-8 \cdot \frac{1}{2} + 7} \le 0$

$\frac{2}{ \frac{1}{4} -4 + 7} \le 0$

$\frac{2}{ \frac{13}{4} } \le 0$

$\frac{8}{ 13} \le 0$

False; exclude .

: Set :

$\frac{4x}{x^{2}-8x + 7} \le 0$

$\frac{4(2)}{2^{2}-8(2) + 7} \le 0$

$\frac{8}{4-16 + 7} \le 0$

$\frac{8}{-5} \le 0$

$-\frac{8}{5} \le 0$

True; include

$(7, \infty)$ : Set :

$\frac{4x}{x^{2}-8x + 7} \le 0$

$\frac{4 \cdot 8 }{8^{2}-8 \cdot 8 + 7} \le 0$

$\frac{32}{64-64 + 7} \le 0$

$\frac{32}{ 7} \le 0$

False; exclude $(7, \infty)$ .

Since the inequality symbol is the "is greater than or equal to" symbol, include the zero, 0, of the numerator - but not the other two boundaries, the zeroes of the denominator. This makes the solution set

$(-\infty, 0] \cup (1,7)$ .

Rational Inequalities

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