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# Polynomial and Rational Inequalities

## Roots, asymptotes, intervals, and test points used to find solution sets.

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Practice Polynomial and Rational Inequalities
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Rational Inequalities

Rational Inequalities

There is one step added to the process of solving rational inequalities because a rational function can also change signs at its vertical asymptotes or at a break in the graph. For instance, look at the graph of the function $r(x)=\frac{x}{x^{2}-9}$ below.

If we want to solve the inequality $\frac{x}{x^{2}-9}>0$ , then we need to use the following critical points: $x=0, x=3,$ and $x=-3$ . $x=0$ is the solution of setting the numerator equal to 0, and this gives us the only root of the function. $x=\pm3$ are the vertical asymptotes, the $x-$ coordinates that make the function undefined because putting in 3 or -3 for $x$ will cause a division by zero.

Testing the intervals between each critical point to see if the values in that interval satisfy the function gives us:

Interval Test Point Positive/Negative? Part of Solution set?
$(-\infty,-3)$ -4 - no
(-3, 0) -2 + yes
(0, 3) 2 - no
$(3,+\infty)$ 4 + yes

Thus, the solutions to $\frac{x}{x^{2}-9}>0$ are $x\in(-3,0)\cup(3,+\infty)$ .

Guided Practice

Questions

1) $f(x) \geq \frac{2x+5}{x-1}$

2) $f(x) \leq \frac{x+2}{x^{2}+1}$

Solutions

1) To identify the graph of the inequality $f(x) \geq \frac{2x+5}{x-1}$ , first treat it as if it were the equality $f(x) \geq \frac{2x+5}{x-1}$

For $f(x) \geq \frac{2x+5}{x-1} :$
To find the critical points, identify the value(s) which make the denominator = 0: $x\ne 1$
That gives us a vertical asymptote of $x=1$
The horizontal asymptote becomes apparent as x becomes truly huge and the "+5" and "-1" no longer matter. At that point, we have $f(x) = \frac{2x}{x} \to f(x) = 2$ So the horizontal asymptote is $y=2$
Now that you know the shape of the graph, simply shade the area above the lines, since the original function was f ( x ) is greater-than function, and leave the lines solid since it was a greater-than or equal to .
The final graph should look like:

2) To graph $f(x) \leq \frac{x+2}{x^{2}+1}$ first treat it as if it were $f(x) = \frac{x+2}{x^{2}+1}$

To identify domain limitations, find value(s) which make the denominator = 0: In this case, where $x^2$ , the only variable in the denominator, is added to 0 , any value for x will be positive. So the domain is all real numbers.
With no limitations on the domain, there are no vertical asymptotes.
The horizontal asymptote: $y=0$ becomes apparent as x becomes huge and the "+2" and "+1" no longer have an effect, giving: $f(x) = \frac{x}{x^2} \to f(x) = \frac{1}{x} \to f(x) = 0$ So the horizontal asymptote is 0
Now that you know the shape of the graph, simply shade the area below the lines, since the original function was f ( x ) is less-than function, and leave the lines solid since it was or equal to .
The final graph should look like: