Question

How would you solve this inequality with a less weird method than I used?

x/(x-1) > 0

Answer 1

@agent0smith im sure your method is absolutely fine. YOUR A TEACHER

Answer 2

Curious to know what the weird method is.

Answer 3

\[\Large \frac{ x }{ 1-x } > 0\]

this was how i did so, after first just doing it analytically to solve it.

I didn't want to multiply by 1-x, since we don't know if 1-x is negative and will change the sign. So multiply by 1-x, then multiply it by 1-x again, cos then the sign will remain unchanged

\[\Large \frac{ x }{ 1-x } *(1-x)^2 > 0*(1-x)^2\]
\[\Large x(1-x) > 0\]I mean, it works, i just felt a little dirty.

Answer 4

That's what i would do.

Answer 5

I guess i've just never needed to use that kinda method, something else usually worked. Maybe just cos i've never needed to solve many rational inequalities.

Answer 6

So from there, @agent0smith, how would you proceed with your solution. The solution is usually expressed using interval notation.

Answer 7

That was the easy part (and i had already solved it mentally by finding when it's positive by plugging in values). It's faster to solve by finding the zeroes and finding where x(1-x) is positive vs negative, but this way seems more technical


x > 0 and 1-x > 0 (for x(1-x) > 0 to be positive since both must be positive)
gives x>0 and x<1 or 0 < x < 1 or (0, 1)

or x < 0 and 1-x < 0 (both must be negative for it to be positive
gives x< 0 and x > 1 (which isn't possible).



The way i did it mentally was just to plug in positive values of x >1, and negative values of x, and seeing that the whole expression is negative in both those cases, and positive when 0 < x <1.

Answer 8

I graphed it:

https://www.desmos.com/calculator/afnkzljlkd

Answer 9

Yeah i'd done it on wolfram to see if it had anything useful http://www.wolframalpha.com/input/?i=x%2F%281-x%29+%3E+0

Answer 10

Would you have used the same method if you had to solve something like

\[\frac{3x + 5}{x + 1} >0\]

Answer 11

Hmm, i considered whether this method would work in all situations of this type. btw it came from finding the domain of log(x/(1-x)). But it seems like this method will often work in similar situations, as you don't have to worry about multiplying both sides by a negative with the inequality sign there.

I guess you can just go ahead and use the fact that the numerator and denominator have to be both positive, or both negative for the expression to be positive... works out the same way actually, same method in a way.

Answer 12

I think I may have seen the method before. Seems convenient to use. But what about those that have slant asymptotes?

Answer 13

Seems like you could do something similar... it just gets more complex because then you can have factors in the denominator too, that can make things awkward.

this \[\Large \frac{x^2 - 25}{x + 1} >0\]same thing, both num/denom will have to be both positive or both negative, you just have to account for the fact that the numerator factors. But just set num. and denom. both > 0 and both < 0 and solve.

Answer 14

I notice you are only considering cases where the expression is greater than zero. Suppose the expression is less than zero. How would that change your approach?

Answer 15

Same basic approach for it to be negative;
numerator must be >0 and denom < 0
or
numerator <0 or denom >0

Answer 16

You would have to do a lot of testing it seems.

Answer 17

If you had something like

\[\frac{x^2 - 7x + 4}{2x + 3}<0\]
For example.

I figured that maybe your methods would make things simpler

Answer 18

Yeah, i don't think there's a quick method for these (even with a quadratic, it's faster to do point tests i guess). Like for that, a point test once you know the zeroes of the numerator would speed it up a bit.

Answer 19

Or just graph it and see "okay it's above the x-axis here and here and not here" once you know the zeroes.

Answer 20

That, to me, is the easiest method.

Answer 21

^yeah. It's much easier. It's faster, esp if you can visualize the graph... it's not hard to picture a +x^2 in your head, and decide where it's above the axis, vs a -x^2, since you know where it's above and below if you know the zeroes.

Answer 22

Woah. Okay there genius. Calm it down Einstein :P

Answer 23

Hey, I was stalking around your profile. Yes, I do have a very good method to solve this. Are you still interested?

Answer 24

It is called the wavy-curve method and is based on your idea of the positive-negative thing.

Answer 25

Sure go ahead @ParthKohli

From looking it up, it looks somewhat familiar though. http://kushagrabasti.blogspot.com/2012/06/wavy-curve-method-wavy-curve-method-is.html

Answer 26

^it's like how I would do it, except I would just graph the function on a calculator rather than draw it out.

Answer 27

It is much more convenient to draw a straight line with two points...