Solve Calculus Question: Positive and Negative Cases for x > 1/3

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SUMMARY

The discussion focuses on solving the inequality 1/x < 3 for both positive and negative values of x. For positive x, the solution is straightforward: x > 1/3. For negative x, the inequality must be manipulated carefully; multiplying by a negative number reverses the inequality, leading to the conclusion that x < 1/3. The final solution set includes all x > 1/3 and all x < 0. The key takeaway is the importance of recognizing the sign of the number when performing operations on inequalities.

PREREQUISITES
  • Understanding of basic algebraic inequalities
  • Knowledge of the properties of inequalities when multiplying/dividing by negative numbers
  • Familiarity with solving rational inequalities
  • Basic calculus concepts related to limits and continuity
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  • Study the properties of inequalities in depth, focusing on multiplication and division by negative numbers
  • Practice solving rational inequalities with varying conditions on x
  • Explore the concept of limits in calculus to understand behavior near critical points
  • Learn about the graphical representation of inequalities to visualize solution sets
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Students in introductory calculus courses, educators teaching algebra and calculus concepts, and anyone looking to strengthen their understanding of inequalities and their properties.

shekki510
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we just started our calc class, and our teacher gave us this question
im sure it is really easy, and i don't know why i just can't get it analytically
the question is solve

1
- < 3 solve for positive and negative case.
x

the positive value is easy, X> 1/3 . How would you solve it for negative vaulues analytically?
 
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Multiplying an inequality by a positive number does not change the direction of the inequality. That's why you can say: For x positive, if 1/x< 3 then 1< 3x and 1/3< x.
(I first multiplied by x and then by 1/3, both positive numbers.)

Multiplying an inequality by a negative number reverses the direction of the inequality: For x negative, if 1/x< 3, then 1> 3x (multiply by the negative x and reverse the inequality. Now multiply by the positive number 1/3 and get 1/3> x.
BUT this was assuming x< 0. Since any x< 0 is necessarily less than 1/3, the solution set is: All x> 1/3 and all x< 0.
 
hmm that clears up some stuff i guess but I am not quite sure yet

so the problem is 1/x<3. multiply -1 to both sides, and you get -1/x > -3 (signs are switched) then to solve for X... -1>-3x then divide by -3 (signs are switched?) so x>1/3... i don't get how to get it algebraically. am i just supposed to accept that when u multiply by a negative number, u just swtich the signs and solve?
 
shekki510 said:
hmm that clears up some stuff i guess but I am not quite sure yet

so the problem is 1/x<3. multiply -1 to both sides, and you get -1/x > -3 (signs are switched) then to solve for X... -1>-3x then divide by -3 (signs are switched?) so x>1/3... i don't get how to get it algebraically. am i just supposed to accept that when u multiply by a negative number, u just swtich the signs and solve?

I think you're making it too complicated. Look at it like this:

Given: 1/x < 3, with x < 0.

Now, multiply both sides by x, remembering that when you multiply an inequality by a negative quantity, you change the sense of the inequality. So, you have:

1 > 3x

Now, divide by the positive 3 and you get: (you don't have to change the sense of the inequality now remember)

1/3 > x

You just need to keep track of whether you are multiplying/dividing by a negative or positive entity. Other than that, just solve like a regular algebraic equation.
 
shekki510 said:
hmm that clears up some stuff i guess but I am not quite sure yet

so the problem is 1/x<3. multiply -1 to both sides, and you get -1/x > -3 (signs are switched) then to solve for X... -1>-3x then divide by -3 (signs are switched?) so x>1/3... i don't get how to get it algebraically. am i just supposed to accept that when u multiply by a negative number, u just swtich the signs and solve?

No, I didn't say anything about multiplying by -1. I said multiply by x, in order to get it out of the denominator. To do that you have to consider whether x is positive or negative, just like your teacher told you.

And, no, you are not "just supposed to accept that when u multiply by a negative number, u just swtich the signs and solve?". You are supposed to have learned that when you learned arithmetic:

3< 5. How do -3 and -5 compare?
 

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