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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:

 

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