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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)=xx29\begin{align*}r(x)=\frac{x}{x^{2}-9}\end{align*} below.

If we want to solve the inequality xx29>0\begin{align*}\frac{x}{x^{2}-9}>0\end{align*}, then we need to use the following critical points: x=0,x=3,\begin{align*}x=0, x=3,\end{align*} and x=3\begin{align*}x=-3\end{align*}. x=0\begin{align*}x=0\end{align*} is the solution of setting the numerator equal to 0, and this gives us the only root of the function. x=±3\begin{align*}x=\pm3\end{align*} are the vertical asymptotes, the x\begin{align*}x-\end{align*}coordinates that make the function undefined because putting in 3 or -3 for x\begin{align*}x\end{align*} 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?
(,3)\begin{align*}(-\infty,-3)\end{align*} -4 - no
(-3, 0) -2 + yes
(0, 3) 2 - no
(3,+)\begin{align*}(3,+\infty)\end{align*} 4 + yes

Thus, the solutions to xx29>0\begin{align*}\frac{x}{x^{2}-9}>0\end{align*} are x(3,0)(3,+)\begin{align*}x\in(-3,0)\cup(3,+\infty)\end{align*}.

Guided Practice

Questions

1) f(x)2x+5x1\begin{align*}f(x) \geq \frac{2x+5}{x-1}\end{align*}

2) f(x)x+2x2+1\begin{align*}f(x) \leq \frac{x+2}{x^{2}+1}\end{align*}

Solutions

1) To identify the graph of the inequality f(x)2x+5x1\begin{align*}f(x) \geq \frac{2x+5}{x-1}\end{align*}, first treat it as if it were the equality f(x)2x+5x1\begin{align*}f(x) \geq \frac{2x+5}{x-1}\end{align*}

For f(x)2x+5x1:\begin{align*}f(x) \geq \frac{2x+5}{x-1} :\end{align*}
To find the critical points, identify the value(s) which make the denominator = 0: x1\begin{align*}x\ne 1\end{align*}
That gives us a vertical asymptote of x=1\begin{align*}x=1\end{align*}
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)=2xxf(x)=2\begin{align*}f(x) = \frac{2x}{x} \to f(x) = 2\end{align*} So the horizontal asymptote is y=2\begin{align*}y=2\end{align*}
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)x+2x2+1\begin{align*}f(x) \leq \frac{x+2}{x^{2}+1}\end{align*} first treat it as if it were f(x)=x+2x2+1\begin{align*}f(x) = \frac{x+2}{x^{2}+1}\end{align*}

To identify domain limitations, find value(s) which make the denominator = 0: In this case, where x2\begin{align*}x^2\end{align*}, 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\begin{align*}y=0\end{align*} becomes apparent as x becomes huge and the "+2" and "+1" no longer have an effect, giving: f(x)=xx2f(x)=1xf(x)=0\begin{align*}f(x) = \frac{x}{x^2} \to f(x) = \frac{1}{x} \to f(x) = 0\end{align*} 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: