# Stupid question for calculus

1. Sep 2, 2004

### shekki510

we just started our calc class, and our teacher gave us this question
im sure it is really easy, and i dont know why i just cant 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?

2. Sep 2, 2004

### HallsofIvy

Staff Emeritus
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.

3. Sep 2, 2004

### shekki510

hmm that clears up some stuff i guess but im 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 dont 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?

4. Sep 2, 2004

### geometer

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.

5. Sep 3, 2004

### HallsofIvy

Staff Emeritus
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?