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Express the set as a union of intervals:

WannaBe

New member
Oct 4, 2013
11
Express the set {X E R: (x+3) (7-x) ((x-2)^2) > 0} as a union of intervals
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
What have you tried ? how to solve that inequality ?
 

WannaBe

New member
Oct 4, 2013
11

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Remember that

\(\displaystyle ab>0 \,\,\, \text{iff }\,\,\,\, a>0,b>0 \,\,\, \text{or}\,\,\, a<0,b<0\)
 

WannaBe

New member
Oct 4, 2013
11

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Note that if you use numbers greater than 7 the inequality will not hold.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I have moved this topic. Although this question does involve sets, rewriting a solution set for an inequality given in set-builder notation to interval notation is a topic typically studied by students of "elementary" algebra.

I don't want to trample on the help being given by ZaidAlyafey, so I will walk you through a similar problem, the way I was taught.

Suppose we are given the set:

\(\displaystyle \{x|(x+4)(x-1)(x-5)^2(x-12)^3>0\}\)

Step 1: Draw a real number line and mark the roots of the polynomial expression:

wannabe1.jpg

Step 2: Consider whether the inequality is weak or strict. If weak, put solid dots at the roots to show they are part of the solution set (giving closed intervals), and if strict put hollow dots to indicate they are not part of the solution (giving open intervals). For this problem, we have a strict inequality so we will put hollow dots at the roots:

wannabe2.jpg

Step 3: Choose test values from within each interval into which the roots have divided the real number line. I will choose some and put them in red, however the choice of values is up to the person working the problem, as long as they are within the intervals:

wannabe3.jpg

Step 4: Put each test value into each factor of the polynomial, and record the resulting signs of each factor, and the consider what the sign the resulting product must be. An even number of negatives gives a positive while an odd number of negatives gives a negative. Take care to make sure the factors with exponents are counted the correct number of times:

wannabe4.jpg

As you become more proficient at this step, you will see that when a root has an odd multiplicity, the sign of the polynomial will change across the root, and when the root has an even multiplicity it will not. The multiplicity of a root refers to how many times it occurs, as indicated by its exponent. Notice the root $x=5$ is of multiplicity 2, and the sign of the polynomial did not change, whereas the root $x=12$ is of multiplicity 3 and the sign did change, as it did also for the other roots of multiplicity 1. Using this information, it is then really only necessary to check one interval, and then apply the information regarding the multiplicity of each root accordingly.

Step 5: Since we are interested in those intervals where the polynomial is positive, we then shade those intervals which resulted in a positive sign:

wannabe5.jpg

Step 6: Write the solution in interval notation:

\(\displaystyle (-4,1)\,\cup\,(12,\infty)\)