You are Here: Home >< Maths

# How to solve inequality with variable with negative power?

Announcements Posted on
Would YOU be put off a uni with a high crime rate? First 50 to have their say get a £5 Amazon voucher! 27-10-2016
1. Hello

I have a derivative (of a function) which I'm using to find out at what values of x the original function is increasing or decreasing. I understand the process. But the derivative is this:

(-2) / (x^3) > 0

(I hope that's written clearly.) I can divide both sides by -1, which also changes the direction of the inequality sign. I can then divide both sides by 2, which leaves me with:

(1)/(x^3) < 0

But I CAN'T divide or multiply both sides by x, because I don't know if the variable is positive or negative. So how do I solve for x? How can I take x out of the denominator? How do I find what x is greater or less than?

This is a question from A-Level Math Core 2. Perhaps I'm misunderstanding the question?

EDIT: Because I'm only solving for what value x is higher or lower than, perhaps in this case I can multiply/divide by a variable, because I will be using both x> and x< to determine when the original function is increasing/decreasing. Am I correct? (I hope I'm explaining myself well.)
2. Hey I've moved this from maths exams to the main maths study help forum.

If 1/(x^3)<0, then we need x^3 to be negative.

If it was positive, lets say for example x=1, we would have 1/1<0 which cannot be true!

So if x^3 is negative, what does this tell us about x?

Does this answer your question? If not, you may need to post the full question you are trying to answer.
3. (Original post by the81kid)
...
You kinda need to just use inspection here. 1/x^3 is going to be negative if and only if x^3 is going to be negative.

x^3 is going to be negative if and only if x is going to be negative.

x is going to be negative is the same thing as x < 0.

And this makes sense, if x < 0, then x^3 < 0. Then 2/x^3 < 0. Then -2/x^3 > 0.
4. (Original post by rayquaza17)
Hey I've moved this from maths exams to the main maths study help forum.

If 1/(x^3)<0, then we need x^3 to be negative.

If it was positive, lets say for example x=1, we would have 1/1<0 which cannot be true!

So if x^3 is negative, what does this tell us about x?

Does this answer your question? If not, you may need to post the full question you are trying to answer.
Hello

Thanks for the reply. The answer section in the book just says (and I'm quoting):

dy/dx = (-2) / (x^3), so function is increasing when x<0 and decreasing when x>0

It doesn't give any steps to find the answer. So I don't know how they got to that answer. I suppose it's something really obvious. I'm supposed to get it by trial and error? In the A-Level Maths I'm doing (Core 1-4, Statistics 1 and Mechanics 1) I don't think they teach us anywhere how to solve this type of equation.
5. (Original post by Zacken)
You kinda need to just use inspection here. 1/x^3 is going to be negative if and only if x^3 is going to be negative.

x^3 is going to be negative if and only if x is going to be negative.

x is going to be negative is the same thing as x < 0.

And this makes sense, if x < 0, then x^3 < 0. Then 2/x^3 < 0. Then -2/x^3 > 0.
Hello
Thanks. So, I'm supposed to just get the answer by trial and error?
I suppose that makes sense.
6. (Original post by the81kid)

dy/dx = (-2) / (x^3), so function is increasing when x > 0.5 and decreasing when x < 0.5
Huh? That's not correct.
7. (Original post by Zacken)
Huh? That's not correct.
That's what I'm thinking...

I think your book is wrong, the81kid.
8. (Original post by Zacken)
Huh? That's not correct.
Ah I'm sorry. I'm tired : / and I copied half the answer from the previous question by mistake. Should read:

dy/dx = (-2) / (x^3), so function is increasing when x<0 and decreasing when x>0
9. (Original post by rayquaza17)
That's what I'm thinking...

I think your book is wrong, the81kid.
No, I'm wrong. I'm really sleepy. Couldn't even read straight when I was copying(!). Sorry about that. I've corrected it now.

dy/dx = (-2) / (x^3), so function is increasing when x<0 and decreasing when x>0
10. (Original post by the81kid)
Ah I'm sorry. I'm tired : / and I copied half the answer from the previous question by mistake. Should read:

dy/dx = (-2) / (x^3), so function is increasing when x<0 and decreasing when x>0
Yeah, now that makes sense.

So -2/x^3 is positive when x^3 is also negative. Then you get negative/negative = positive. But x^3 is only negative when x is negative since negative^3 = negative. So -2/x^3 > 0 when x < 0. (Hence increasing there)

-2/x^3 is negative when x^3 is positive. Then you get negative/positive = negative. But x^3 is only positive when x is positive since positive^3 = positive. So -2/x^3 < 0 when x > 0. (Hence decreasing there)
11. (Original post by Zacken)
Yeah, now that makes sense.

So -2/x^3 is positive when x^3 is also negative. Then you get negative/negative = positive. But x^3 is only negative when x is negative since negative^3 = negative. So -2/x^3 > 0 when x < 0. (Hence increasing there)

-2/x^3 is negative when x^3 is positive. Then you get negative/positive = negative. But x^3 is only positive when x is positive since positive^3 = positive. So -2/x^3 < 0 when x > 0. (Hence decreasing there)
Thanks. That makes sense. So I was supposed to just use common sense? I was thinking that I could, but then I thought there are some steps I should show I've used. I understand now. Thanks a lot, again!
12. (Original post by the81kid)
Thanks. That makes sense. So I was supposed to just use common sense? I was thinking that I could, but then I thought there are some steps I should show I've used. I understand now. Thanks a lot, again!
Nah, there's no working needed. This is more of a "state where the function is increasing and decreasing" question instead of a "find where the function is increasing or decreasing" question. The latter means you need to do some work, the former means just using common sense.
13. (Original post by Zacken)
Nah, there's no working needed. This is more of a "state where the function is increasing and decreasing" question instead of a "find where the function is increasing or decreasing" question. The latter means you need to do some work, the former means just using common sense.
Sounds good. I feel happier now. Thought I'd missed something.
14. (Original post by the81kid)
Sounds good. I feel happier now. Thought I'd missed something.
Awesome. :-)

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: April 15, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

Find out here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams