How do you solve rational inequalities?

Question

4626 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-10-31T16:05:16

Let us solve the following rational inequality.

$f (x) = \frac{x + 1}{x^{2} + x - 6} \leq 0$ $f(x)={x+1}/{x^2+x-6} le 0$

Set the numerator equal to zero, and solve for $x$ $x$ .

$x + 1 = 0 \Rightarrow x = - 1$ $x+1=0 => x=-1$

(Note: $f (- 1) = 0$ $f(-1)=0$ )

Set the denominator equal to zero, and solve for $x$ $x$ .

$x^{2} + x - 6 = (x + 3) (x - 2) = 0 \Rightarrow x = - 3, 2$ $x^2+x-6=(x+3)(x-2)=0 => x=-3,2$

(Note: $f (- 3)$ $f(-3)$ and $f (2)$ $f(2)$ are undefined.)

Using $x = - 3, - 1, 2$ $x=-3,-1,2$ above to split the number line into open intervals:

$(- \infty, - 3), (- 3, - 1), (- 1, 2)$ $(-infty,-3), (-3,-1),(-1,2)$ , and $(2, \infty)$ $(2,infty)$

Using sample numbers $x = - 4, - 2, 0, 3$ $x=-4,-2,0,3$ for each interval above, respectively, we can determine the sign of (LHS).

$f (- 4) = - 2 < 0 \Rightarrow f (x) < 0$ $f(-4)=-2<0 => f(x)<0$ on $(- \infty, - 3)$ $(-infty,-3)$

$f (- 2) = \frac{1}{4} > 0 \Rightarrow f (x) > 0$ $f(-2)=1/4>0 => f(x)>0$ on $(- 3, - 1)$ $(-3,-1)$

$f (0) = - \frac{1}{6} < 0 \Rightarrow f (x) < 0$ $f(0)=-1/6<0 => f(x)<0$ on $(- 1, 2)$ $(-1,2)$

$f (3) = \frac{2}{3} > 0 \Rightarrow f (x) > 0$ $f(3)=2/3>0 => f(x)>0$ on $(2, \infty)$ $(2,infty)$

Hence, $f (x) \leq 0$ $f(x) le 0$ on $(- \infty, - 3) \cup [- 1, 2)$ $(-infty,-3)cup[-1,2)$ .

(Note: $- 1$ $-1$ is included since $f (- 1) = 0$ $f(-1)=0$ .)

The graph of $y = f (x)$ $y=f(x)$ looks like:

enter image source here

I hope that this was helpful.