6.6 Solving Polynomial And Rational Inequalities

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Example 1

Solve: x(x+3)2(x−4)<0.

Solution:

Begin by finding the critical numbers. For a polynomial inequality in standard form, with zero on one side, the critical numbers are the roots. Because f(x)=x(x+3)2(x−4) is given in its factored form the roots are apparent. Here the roots are: 0, −3, and 4. Because of the strict inequality, plot them using open dots on a number line.

In this case, the critical numbers partition the number line into four regions. Test values in each region to determine if f is positive or negative. Here we choose test values −5, −1, 2, and 6. Remember that we are only concerned with the sign (+ or −) of the result.

f(−5)=(−5)(−5+3)2(−5−4)=(−)(−)2(−)=+Positivef(−1)=(−1)(−1+3)2(−1−4)=(−)(+)2(−)=+Positivef(2)=(2)(2+3)2(2−4)=(+)(+)2(−)=−Negativef(6)=(6)(6+3)2(6−4)=(+)(+)2(+)=+Positive

After testing values we can complete a sign chart.

The question asks us to find the values where f(x)<0, or where the function is negative. From the sign chart we can see that the function is negative for x-values in between 0 and 4.

We can express this solution set in two ways:

{x|0<x<4}Set notation(0,4)Interval notation

In this textbook we will continue to present solution sets using interval notation.

Answer: (0,4)

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