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Solution of Rational Inequalities by Graphing

In document College Algebra (Page 103-111)

In the previous section, we saw how to solve polynomial inequalities by graph- ing. In this section, we will use similar methods to solve rational inequalities. Rational inequalities involve ratios of polynomials or fractions. Because these types of problems involve fractions, the graphs of the functions that we work with will have what are known as asymptotes. This word comes from a Greek root having to do with two lines that come very close to each other but never meet.

The vertical asymptotes of a graph will appear at places where the original ex- pression has a zero denominator. This means that the function is not defined at thosexvalues and so, rather than having ayvalue at that point, the graph has an asymptote.

Example

Below is a graph of the functiony= x+ 2

x−1

−8 −6 −4 −2 2 4 6 8

−5 5

Rather than having ay value at the point wherex = 1, the dotted line indicates the asymptote where the function is not defined. In the previous section, we were interested in finding the roots of the function because these are the places where y= 0, and can be the dividing points between where theyvalues are greater than zero (y >0) and the where theyvalues are less than zero (y <0).

2.4. SOLUTION OF RATIONAL INEQUALITIES BY GRAPHING 101 The importance of the asymptotes in analyzing rational functions is that, like the roots, these represent x values that can be the dividing points between where y >0and wherey <0.

Example

Solve the given inequality.

2

x−3 >0

First we examine the graph:

−8 −6 −4 −2 2 4 6 8

−5 5

Notice that the asymptote for this graph occurs at the valuex= 3, because this is thexvalue that creates a zero denominator. Also notice that theyvalues switch from being negative to being positive across the asymptote.

There are no roots for this function because there are noxvalues that makey= 0. For a fraction to be zero, the numerator must equal zero. In this example the numerator is 2 and no value of x will make it equal zero. Therefore the only possible dividing point on the graph isx = 3, and the solution to the inequality isx >3.

Example

Solve the given inequality. x−2

x−3 >0

−8 −6 −4 −2 2 4 6 8

−5 5

In this inequality, there is again an asymptote atx = 3, but there is also a root at the value x = 2, because whenx = 2. y = 2−2

2−3 = 0

−1 = 0. So we have two

dividing points to consider, x = 2 and x = 3. We can see from the graph that y >0forx <2orx >3, so that is the solution to the given inequality.

Example

Solve the given inequality. x2 −2

x−3 >0

In this problem, we have the same asymptote as the previous two problems:x= 3. However, in this inequality, there are two roots, because there are twoxvalues that make the numerator equal zero.

x2−2 = 0means thatx2 = 2andx=±√2≈ ±1.414

2.4. SOLUTION OF RATIONAL INEQUALITIES BY GRAPHING 103 −8 −6 −4 −2 2 4 6 8 −5 5 10 15

In the graph above, we can see the asymptote at x = 3 and the two roots at x≈1.414,−1.414.

Thexvalues that makey >0are−1.414< x < 1.414ORx >3.

Example

Solve the given inequality. x−2

x23 >0

−8 −6 −4 −2 2 4 6 8

−5 5

The roots for this function are thexvalues that make the numerator equal zero: x−2 = 0, thereforex= 2, and we can see this root on the graph.

The asymptotes for the function are thexvalues that make the denominator equal zero:

x2−3 = 0means thatx2 = 3andx=±√3≈ ±1.732

−8 −6 −4 −2 2 4 6 8

−5 5

Therefore the solution for the given inequality is:

−1.732 < x <1.732ORx >2 Example

Solve the given inequality.

5x+ 1

x2 + 3x4 <0

−8 −6 −4 −2 2 4 6 8

−5 5

2.4. SOLUTION OF RATIONAL INEQUALITIES BY GRAPHING 105 Roots 5x+ 1 = 0 5x=−1 x=−0.2 =−1 5 Asymptotes x2+ 3x4 = 0 (x+ 4)(x−1) = 0 x=−4,1 −8 −6 −4 −2 2 4 6 8 −5 5

If we combine the algebraic analysis above with what we see in the graph, then we know that the dividing points important to the solution of this inequality are atx=−4,−0.2,1. The intervals where theyvalues are less than zero arex <−4

Example x2 + 2x−1 x2+ 7x+ 5 ≤0 −8 −6 −4 −2 2 4 6 8 −5 5 Roots x2+ 2x−1 = 0 x≈ −2.414,0.414 Asymptotes x2+ 7x+ 5 = 0 x≈ −6.193,−0.807

We can see that the dividing points important to the solution of the inequality are x ≈ −6.193,−2.414,−0.807,0.414. The intervals where the yvalues are less than or equal to zero are−6.193 ≤x≤ −2.414OR−0.807≤x≤0.414.

2.4. SOLUTION OF RATIONAL INEQUALITIES BY GRAPHING 107

Exercises 2.4

Solve each inequality.

1) x+ 4 x28x+ 12 >0 2) 2x+ 3 x22x35 <0 3) x 25x14 x2+ 3x10 <0 4) 2x2x3 x2+ 10x+ 16 >0 5) 3x+ 2 x2+x5 <0 6) x2+ 2x+ 5 x23x7 >0 7) x 3+ 9 x2+x1 >0 8) x3+ 9 x2+x+ 1 >0

Solve each inequality.

9) x 22x9 3x+ 11 >0 10) x2+ 4x+ 3 2x+ 1 <0 11) x 2+x5 x2x6 >0 12) x3+ 2 x22 >0 13) x 2+ 2x7 x2+ 3x6 <0 14) 2x−x2 x24x+ 6 <0 15) x 27 x2+ 5x1 <0 16) x−5 3x22x3 >0

In document College Algebra (Page 103-111)

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