Rational Inequality 2

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? \(\;\) Only you can say what you "must" do. (There are different methods.)

In general, I factor what I can when working with an algebraic ratio, to look for simplifications (eg: common factors that can be canceled). In this exericse, if we remember the factorization for a difference of cubes, then we can see there are no common factors above.

Once I know the ratio is in lowest terms, I may or may not continue with a completely-factored form. It depends on the method I choose for the exercise. Since you asked about the first step, it seems like you haven't decided on a method. (Did you review in your book first, or is this something you copied from another site?)

One could construct a sign chart, to determine where the ratio is positive and where it is negative. Using that method, I would leave the given ratio just as it is. I would find the zeros of the numerator and denominator, to divide the Real number line into intervals. Within each interval, I would pick an easy test value and determine the sign of each part (i.e., the sign of each of the two parts in the numerator and the sign of the denominator). Organizing the signs in a chart allows us to determine the sign of the ratio itself (within each interval).

For example, if one of the intervals is (1,3), then we could pick 2 as an easy test value. For x=2, we get the signs for each part:

(3 - x)^3 is positive
(2x + 1) is positive
(x^3-1) is positive

So, the given ratio is: (positive × positive)/positive, and that's a positive number. Because the ratio is greater than zero, the interval (1,3) is not part of the solution.

If you're interested in constructing a sign chart, then check your book's index or google for videos/lessons (to review, first). Otherwise, please show what you can try (maybe another method) or explain why you're stuck. Cheers

?
 
Otis said:
? \(\;\) Only you can say what you "must" do. (There are different methods.)

In general, I factor what I can when working with an algebraic ratio, to look for simplifications (eg: common factors that can be canceled). In this exericse, if we remember the factorization for a difference of cubes, then we can see there are no common factors above.

Once I know the ratio is in lowest terms, I may or may not continue with a completely-factored form. It depends on the method I choose for the exercise. Since you asked about the first step, it seems like you haven't decided on a method. (Did you review in your book first, or is this something you copied from another site?)

One could construct a sign chart, to determine where the ratio is positive and where it is negative. Using that method, I would leave the given ratio just as it is. I would find the zeros of the numerator and denominator, to divide the Real number line into intervals. Within each interval, I would pick an easy test value and determine the sign of each part (i.e., the sign of each of the two parts in the numerator and the sign of the denominator). Organizing the signs in a chart allows us to determine the sign of the ratio itself (within each interval).

For example, if one of the intervals is (1,3), then we could pick 2 as an easy test value. For x=2, we get the signs for each part:

(3 - x)^3 is positive
(2x + 1) is positive
(x^3-1) is positive

So, the given ratio is: (positive × positive)/positive, and that's a positive number. Because the ratio is greater than zero, the interval (1,3) is not part of the solution.

If you're interested in constructing a sign chart, then check your book's index or google for videos/lessons (to review, first). Otherwise, please show what you can try (maybe another method) or explain why you're stuck. Cheers

?
Click to expand...

Solution:

(3 - x)^3(2x+1)/(x^3 - 1) < 0

Multiply both sides by the denominator.

(3 - x)^3(2x+1) < 0

I then solve each factor for x.

Doing so, I get x = -1/2 and x = 3.

Place each x-value on the number line.

<--------(-1/2)---------------(3)--------->


When x = -2, we get
-375 < 0. True statement.

When x = 0, we get
27 < 0. False statement.

When x = 4, we get
-9 < 0. True statement.

Solution: (-00, -1/2) U (3, 00).

Note:

-00 means negative infinity

00 means positive infinity.
 
You say
"(3 - x)^3(2x+1)/(x^3 - 1) < 0

Multiply both sides by the denominator.

(3 - x)^3(2x+1) < 0"
Do you recall that multiplying both sides of an inequality by a negative number reverses the direction of the inequality? You need to consider two different cases: (1) x^3- 1> 0 and (2) x^3- 1< 0.

Personally, I wouldn't do it that way. A product of factors (which includes division) is negative if and only if an odd number of the factors is negative. Since x^3- 1= (x- 1)(x^2+ x+ 1) the factors are 3- x (three times), 2x+ 1, x-1, and x^2+ x+ 1. For what values of x are each of those negative or positive? On what intervals are an odd number of the 6 factors negative? (Completing the square, x^2+ x+ 1= x^2+ x+ 1/4- 1/4+ 1= (x- 1/2)^2+ 3/4 so x^2+ x+ 1 is never negative.)
 
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HallsofIvy said:
You say
"(3 - x)^3(2x+1)/(x^3 - 1) < 0

Multiply both sides by the denominator.

(3 - x)^3(2x+1) < 0"
Do you recall that multiplying both sides of an inequality by a negative number reverses the direction of the inequality? You need to consider two different cases: (1) x^3- 1> 0 and (2) x^3- 1< 0.

Personally, I wouldn't do it that way. A product of factors (which includes division) is negative if and only if an odd number of the factors is negative. Since x^3- 1= (x- 1)(x^2+ x+ 1) the factors are 3- x (three times), 2x+ 1, x-1, and x^2+ x+ 1. For what values of x are each of those negative or positive? On what intervals are an odd number of the 6 factors negative? (Completing the square, x^2+ x+ 1= x^2+ x+ 1/4- 1/4+ 1= (x- 1/2)^2+ 3/4 so x^2+ x+ 1 is never negative.)
Click to expand...

I'm gonna try considering two cases as you pointed out.
 
HallsofIvy said:
You say
"(3 - x)^3(2x+1)/(x^3 - 1) < 0

Multiply both sides by the denominator.

(3 - x)^3(2x+1) < 0"
Do you recall that multiplying both sides of an inequality by a negative number reverses the direction of the inequality? You need to consider two different cases: (1) x^3- 1> 0 and (2) x^3- 1< 0.

Personally, I wouldn't do it that way. A product of factors (which includes division) is negative if and only if an odd number of the factors is negative. Since x^3- 1= (x- 1)(x^2+ x+ 1) the factors are 3- x (three times), 2x+ 1, x-1, and x^2+ x+ 1. For what values of x are each of those negative or positive? On what intervals are an odd number of the 6 factors negative? (Completing the square, x^2+ x+ 1= x^2+ x+ 1/4- 1/4+ 1= (x- 1/2)^2+ 3/4 so x^2+ x+ 1 is never negative.)
Click to expand...



Case (1):

x^3 - 1 > 0

Rewrite as x^3 - 1 = 0. Solve for x.

x = 1

Place on number line.

<----------(1)---------->

When x = 0, we get -1 > 0. False statement.

When x = 2, we get 7 > 0.
True statement.

Solution: (1, 00).


Case (2):

x^3- 1< 0

Solving for x, we get 1.
Back to the number line.

<----------(1)--------->

When x = 0, we get -1 < 0. True statement.

When x = 2, we get 7 < 0. False statement.

Solution: (-00, 1).

Is any of this right?
 
Yes, x^3- 1 is negative for x< 1 and positive for x> 1. You also need to look at where 2x+ 1 is positive or negative and where 2- x is positive and negative to determine the sign of the function.
 
HallsofIvy said:
Yes, x^3- 1 is negative for x< 1 and positive for x> 1. You also need to look at where 2x+ 1 is positive or negative and where 2- x is positive and negative to determine the sign of the function.
Click to expand...

After much work on this problem, I came to the following interval solution:

(-1/2, 1) U (3, 00)
 
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