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Interval Notation


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Posted by dolphin@eskimo.com (64.24.187.30) on August 11, 2001 at 20:21:33:

In Reply to: Inequalities posted by on August 11, 2001 at 17:19:34:


The thing to realize on your first problem is this: the x-intercepts of the expression divide the x-axis into intervals where the expression is either greater than zero or less than zero. To determine where the expression is less than zero, we pick a test value in each interval. If our test value leads to a true statement using the original inequality, then we know that every value in that interval is a solution.

The expression is given in factored form, so it is easy to find the x-intercepts by inspection. The x-intercepts are -4, -1, 0, and 1. These intercepts divide the x-axis into the following five intervals:

(-infinity,-4) (-4,-1) (-1,0) (0,1) (1,infinity)

Now we need to pick a test value in each interval, and check it to see if it leads to a true statement. Let's pick -5 as our test value in the first interval.

(x)(x^2 - 1)(x + 4)^2 < 0

(-5)((-5)^2 - 1)(-5 + 4)^2 < 0

(-5)(25 - 1)(-1)^2 < 0

(-5)(24)(1) < 0

-120 < 0

The test value leads to a true statement, so all of the values in the interval (-infinity,-1) are solutions.

The next interval is (-4,-1). Let's pick -2 to test.

(x)(x^2 - 1)(x + 4)^2 < 0

(-2)((-2)^2 - 1)(-2 + 4)^2 < 0

(-2)(4 - 1)(2)^2 < 0

(-2)(3)(4) < 0

-24 < 0

This is a true statement also, so all of the values in the interval (-4,-1) are solutions.

We process the remaining three intervals the same way.

Interval: (-1,0) Test value: -1/2 Result: 147/32 < 0 False

Interval: (0,1) Test value: 1/2 Result: -243/32 < 0 True

Interval: (1,infinity) Test value: 2 Result: 216 < 0 False

Putting all of this information together gives us our solution.

The solution set is (-infinity,-1) U (0,1)

On your second problem, you should memorize the following:

|expression| < c means -c < expression < c

In other words, you can remove the absolute value symbols by placing the expression inbetween the negative and positive value of the constant.

|2x + 3| < 7

-7 < 2x + 3 < 7

Now, we can subtract three from each quantity and divide by 2 to isolate x:

-7 - 3 < 2x < 7 - 3

-10 < 2x < 4

-5 < x < 2

The solution is (-5,2).

~ Mark



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