Exponents question

[dslewis]

Generic exponents question:

For what values does X^3>X hold true

When I solve this algebraically I get x>1; X>-1, but I also get x>0. THe answer is -1<x<0 and x>1, but why do you change the sign from x>0 to X<0? Can someone please explain this. THanks

[KLM]

Generic exponents question:

For what values does x^3 > x hold true

When I solve this algebraically I get x > 1; X > -1, but I also get x > 0. THe answer is -1 < x < 0 and x > 1, but why do you change the sign from x > 0 to x < 0? Can someone please explain this. Thanks

x^3 > x
x^3 - x > 0
x (x^2 - 1) > 0
x (x - 1) (x + 1) > 0

from above: x > 0, x > 1, and x > - 1.
but if x is a +ve fraction, then x^3 > x doesnot hold true. so x has to be >1 and <0.

so -1 < x < 0. also x>1.

[suyash]

PERFECT EXPLANATION!!!

[dslewis]

What are the exact rules or steps inorder to do this problem?

[dslewis]

yeah but verbally what is the rule cause I cannot see it in your illustration. It seems confusing and I've done these problems a millions different ways but just cant spot the rules.

I get confused because I'm not sure when its the upper limit or the lower limit or if its both. For example :
x^2<x
x<0 x<1
X>0 X>1
Answer: 0<x<1 Why though?

x^2>x
x>0 x>1
x<0 x<1
Answer: x<0 and x>1 why? If I knew the rule I could figure it out. I try all combo's but nothing seems to work.

[David Pollack]

I like the way KLM started this one off:

--------

x^3 > x
x^3 - x > 0
x (x^2 - 1) > 0
x (x - 1) (x + 1) > 0

---------

These 4 statements are certainly equivalent to each other, so let's look at the last one.

I'll show how to solve this by a direct (but slightly subtle) method and then by a partly
trial-and-error approach (maybe the easier way to understand, but harder to actually
carry out)


First the direct method:

The expression x (x-1) (x+1) is the product of 3 numbers. In order for such a product to
be positive, we must either have all 3 terms being positive, or 1 term being positive and 2 terms
being negative.

List the terms in increasing order: x-1, x, x+1

So having all three positive is the same as having the smallest one positive - that is having x-1>0 or equivalently x>1.

Having one term positive and two terms negative can only happen when x-1 and x are negative and x+1 is positive.
So in this case x<0 and x+1>0, so x<0 and x>-1.

Thus the solution is x>1 together with -1<x<0.

---

Now the other way to think about it:

We want to know when x^3-x = x(x-1)(x+1) is positive. The idea is to think about where this
expression might shift from being positive to negative or from being negative to positive.
The only time this can happen is at x values where the expression is 0.
That is, the only times x(x-1)(x+1) can change sign are when x=-1, x=0, or x=1.

So think about a number line with -1, 0, and 1 marked

------------- -1 -------- 0 ----------- 1 --------

The line is broken into four pieces:
i) x<-1
ii) -1<x<0
iii) 0<x<1
iv) x>1

The expression x(x-1)(x+1) stays the same sign on the whole length of each of these intervals. We want
to know on which of these intervals it is positive, and on which it is negative. So let's just check!

For example, the number x=-2 is in the first interval. But when we plug -2 into our expression [either into x^3-x
or into x(x+1)(x-1) works since they are equal after all] we get -6. So the expression is negative at this point and thus
is negative on the entire first interval. So this interval doesn't work.

On the other hand, -1/2 is in the second interval and if we plug -1/2 into our expression we get 3/8 which is positive.
Thus the expression is positive on the whole interval and -1<x<0 should be part of our answer.

Continuing, 1/2 is in the third interval and plugging in gives us -3/8 which negative, so this interval is no good.

Finally 2 is in the fourth interval and plugging in gives 6 which is positive, and so the expression is
positive on the 4th interval and we should include x>1 in our answer.

So the final answer is -1<x<0 or x>1.

[KLM]

Generic exponents question:

For what values does x^3 > x hold true

When I solve this algebraically I get x > 1; X > -1, but I also get x > 0. THe answer is -1 < x < 0 and x > 1, but why do you change the sign from x > 0 to x < 0? Can someone please explain this. Thanks

x^3 > x
x^3 - x > 0
x (x^2 - 1) > 0
x (x - 1) (x + 1) > 0

from above: x > 0, x > 1, and x > - 1.
but if x is a +ve fraction, then x^3 > x doesnot hold true. so x has to be >1 and <0.

so -1 < x < 0. also x>1.

http://gmatclub.com/forum/7-t61602

[rfernandez]

Nice solutions, David.