Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Math Analysis


Date: 09/13/97 at 19:38:19
From: Raymond
Subject: Math Analysis

In the following problems I have to figure out the boundaries for the 
inequality. I get as far as breaking up the inequality but from there 
I can't figure out what step to do next. The problems are as follows:

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

   2)  4x^3-6x^2 is less than or equal to 0.

   3)  4x(x^2-6)/x^2-4 is less than 0.

I appreciate your help!


Date: 09/14/97 at 08:57:03
From: Doctor Anthony
Subject: Re: Math Analysis

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

When the expression is factorized like this the answer is particularly 
easy to write down. We can see that the lefthand side is a cubic 
polynomial, with the term in x^3 being positive. Such a curve has 
a characteristic S shape with the the lefthand end disappearing to 
-infinity for both x and y, the righthand end to + infinity for x 
and y, and in the middle we get a maximum turning point first; then, 
as x increases, a minimum turning point before the curve bends up and 
goes off to +infinity.  

Now this curve cuts the x axis (i.e. making the whole expression = 0) 
at x = -1/3, at x = 0 and at x = 2/3. With this picture in mind, it is 
clear that the curve is negative if x < -1/3, also if x lies between 
0 and 2/3; otherwise it is positive. So the inequality is satisfied 
when:
          x < -1/3   and  0 < x < 2/3

>   2)  4x^3-6x^2 is less than or equal to 0.

Again take factors  2x^2(2x-3) <= 0

This is the same shape as that described before for a cubic 
polynomial, except that the maximum turning point at x = 0 just 
touches the x axis and does not go above it before bending down 
towards the minimum turning point. 

After the minimum point the curve bends up and crosses the x axis 
at x = 3/2.  So again we can picture where the curve lies below the 
x axis, thereby satisfying the condition that it be less than 0.

Required condition is  x <=  3/2

>   3)  4x(x^2-6)/x^2-4 is less than 0.

Although one of the terms in this expression divides the other two, 
we get the same result if we think of it as multiplying the other two.  
This is because you get a change of sign whether you multiply or 
divide by a negative quantity, and it is easy to visualize a curve's 
shape if all the terms multiply. So I am changing this to a different 
problem with all terms multiplying, but the answers will be the same.

  4x(x^2-6)(x^2-4) <= 0    carry out further factorizing as possible:

  4x(x-sqrt^6))(x+sqrt(6))(x-2)(x+2) <= 0

Now the places where the curve cuts the x axis in ascending order are:

  -sqrt(6), -2,  0,  +2,  +sqrt(6)

Now you must imagine the 5th power curve behaving like the cubic 
already described, but with an extra hump. So coming from the left, 
the curve cuts the x axis from below at -sqrt(6). Thus the inequality 
is satisfied with x <= -sqrt(6). We now get a positive region between 
-sqrt(6) and -2. Then a negative region between -2 and 0. So the 
inequality is satisfied with -2 <= x < = 0.  From here the curve 
goes positive again between 0 and +2, but from +2 to +sqrt(6) the 
curve is again below the x axis. So the inequality is satisfied with 
2 <= x <= sqrt(6).

Putting all these facts together we get the inequality satisfied when:

   x <= -sqrt(6),   -2 <= x <= 0,     2 <= x <= sqrt(6)

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Analysis
High School Basic Algebra

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/