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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 r(x)=xx29 below.

If we want to solve the inequality xx29>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=±3 are the vertical asymptotes, the xcoordinates 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?
(,3) -4 - no
(-3, 0) -2 + yes
(0, 3) 2 - no
(3,+) 4 + yes

Thus, the solutions to xx29>0 are x(3,0)(3,+).

 

Guided Practice

Questions

1) f(x)2x+5x1

2) f(x)x+2x2+1

Solutions

 

 

1) To identify the graph of the inequality f(x)2x+5x1, first treat it as if it were the equality f(x)2x+5x1

For f(x)2x+5x1:
To find the critical points, identify the value(s) which make the denominator = 0: x1
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)=2xxf(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)x+2x2+1 first treat it as if it were f(x)=x+2x2+1

To identify domain limitations, find value(s) which make the denominator = 0: In this case, where x2, 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)=xx2f(x)=1xf(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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