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How to solve inequality with variable with negative power?

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.)

(edited 8 years ago)

Reply 1

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. :smile:

Reply 2

Zacken

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.

Reply 3

the81kid

Original post by rayquaza17

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.

(edited 8 years ago)

Reply 4

the81kid

Original post by Zacken

Hello
Thanks. So, I'm supposed to just get the answer by trial and error?
I suppose that makes sense.

Reply 5

Zacken

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.

Reply 6

rayquaza17

Original post by Zacken

Huh? That's not correct.

That's what I'm thinking...

I think your book is wrong, @the81kid.

Reply 7

the81kid

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

Reply 8

the81kid

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.

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

Reply 9

Zacken

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)

Reply 10

the81kid

Original post by Zacken

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!

Reply 11

Zacken

Original post by the81kid

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.