Calculus Methods

02 Solving Polynomial Inequalities

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        1) Get all parts of the inequality on one side of the

        equation, leaving 0 on the other side.

 

 

        2) Gather all terms into a single fraction.

 

 

        3) Factor both the numerator and denominator.

 

        4) Create a "sign chart" to determine when each

        factor is positive and when it is negative.

 

                  A) List all factors in a single column on the

                   left, and add a last row named "P" (for

                   product).

 

                  B) Draw a vertical line for each distinct zero

                   of each factor.

 

                  C) At the bottom of each line, write the

                   zeroes (roots) down in increasing numerical

                   order. These numbers form a number line.

 

                  D) For each row (each factor), write a "+" or

                   "-" to indicate whether the given factor is

                   positive or negative on the intervals between

                   successive zeroes.

 

                  E) For the "P" row, simply multiply all of the

                   +s and -s together to find out whether the

                   total product is positive or negative on that

                   interval.

 

        5) Check the factored inequality that you obtained

        in step (3). Are you looking for the positive

        products (greater than 0) or the negative parts (less

        than 0)? Also, determine when given endpoints

        (zeroes) can be included, such as "less than or equal

        to", and when they cannot, such as "less than" and

        zeroes in the denominator. Infinity is NEVER

        included IN a domain! It is not a number! Write

        down your final interval(s).

 

 

        COMMON PITFALL: Make sure that all terms are

        on one side and 0 is on the other. Since we are

        measuring when factors are positive or negative, we

        need to be comparing the factors to zero.

 

        COMMON PITFALL: Make sure to include an

        overall negative constant, assuming there is one. It

        is always negative in each interval, and that will

         change the overall sign of that interval. To avoid

        having to include this, multiply both sides of your

        final equation by -1 (make sure to change the

        direction of the inequality!) to rid yourself of the

        negative term.

 

        TIP: Factors change sign at their own zeros, so it's

        not necessary to check every single interval for

        every single factor. Just figure out which zero or

        zeroes belong to a particular factor and determine

        the sign on one interval. From there simply make

        sure to change signs at its zero or zeroes.

 

        TIP: Factors that have even powers do not need to

        be included in a sign chart since all of the

        occurrences of that factor will always net a positive

        result.

 

        TIP: Factors that have odd powers only need to be

        listed once (see above tip).

 

 

 

         Example #1: Find all solutions of the inequality: .

 

         1)

 

 

         2) It is already a single fraction (denominator of 1!).

 

 

         3)

 

 

         4A)

 

 

 

 

 

 

         4B)

 

 

 

 

 

 

 

         4C)

 

 

 

 

 

 

 

 

          4D)

 

          4E)

                    

 

         5) We are looking for products that are LESS

        THAN zero, so the answers we care about in

        particular are circled below:

 

         So the two intervals that work are the interval

        between negative infinity and -5 and the interval

        between 7 and infinity. Now we must determine

        which endpoints to include. Since we are dealing

        with "less than or equal to," we should include as

        many endpoints as possible. Since infinity is never

        included, our solution set is thus: .

 

 

 

 

 

         Example #2: Find all solutions of the inequality: .

 

 

         1)

 

 

 

         2)

                 

 

 

 

         3)

                          

 

 

 

         4A) 

 

 

 

 

 

 

 

 

 

 

 

         4B)

 

 

 

 

 

 

 

 

 

 

         4C) The zero of x+1 is -1; the zero of x is 0; the

        zero of x-1 is 1; the zero of x-2 is 2.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

         4D)

 

 

 

 

 

 

 

 

 

 

 

          4E) The are five negatives in the first column, and

        five negatives multiplied together yield a negative

        answer. There are three negatives multiplied by two

        positives in the second column, yielding a negative

        answer again. There are three positives and two

        negatives in the third column, yielding a positive

        answer. &c.

 

 

 

 

 

 

 

 

 

 

 

 

 

         5) We are looking for answers GREATER THAN

        or equal to zero, so the answers we care about in

        particular are circled below:

 

 

 

 

 

 

 

 

 

 

 

 

 

         So the two intervals that work are the interval

        between 0 and 1 and the interval between 2 and

        infinity. Now we must determine which endpoints

        to include. Since we are dealing with "greater than

        or equal to," we should include as many endpoints

        as possible. The ones we need to exclude are 0, 1,

        and 2, since all of these would make the

        denominator equal to zero, which is not allowed.

        Since infinity is never included, our solution set is

        thus: .

 

 

 

 

        Example #3: On the interval , find all

        solutions of the inequality: .

 

         1) Already done.

 

         2) Already done.

 

         3) Already done.

 

 

         4A)

                

 

 

         4B)

 

 

 

 

 

 

 

         4C)

 

 

 

 

 

 

 

 

         4D) Note that since the only interval on which we

        are looking does not include the region less than 0

        or greater than 2, we do not need to fill in those

        regions. Also note that sinx changes sign at its zero,

        , and cosx changes sign at its zeroes, /2 and 3/2.

 

 

 

 

 

 

 

 

 

 

         4E)

 

 

 

 

 

 

 

 

        5) We are looking for answers GREATER THAN

        zero, so the answers we care about in particular are

        circled below:

 

 

 

 

 

 

        

 

 

 

         So the two intervals that work are the interval

        between 0 and /2 and the interval between and

        3/2. Now we must determine which endpoints to

        include. Since we are dealing with "greater than,"

        we do not include any endpoints. Our solution set is

        thus: .

 

 

 

On to Method 03 - Separating Partial Fractions

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