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Polynomial and Rational Inequalities

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

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

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

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

 

Guided Practice

Questions

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

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

Solutions

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

For \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: \begin{align*}x\ne 1\end{align*}
That gives us a vertical asymptote of \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 \begin{align*}f(x) = \frac{2x}{x} \to f(x) = 2\end{align*} So the horizontal asymptote is \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 \begin{align*}f(x) \leq \frac{x+2}{x^{2}+1}\end{align*} first treat it as if it were \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 \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: \begin{align*}y=0\end{align*} becomes apparent as x becomes huge and the "+2" and "+1" no longer have an effect, giving: \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:

 

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