x^4-x<0

How do I solve it?

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- Jul 25th 2011, 03:15 PMRK29Inequality with higher degree
x^4-x<0

How do I solve it? - Jul 25th 2011, 03:21 PMpickslidesRe: Inequality with higher degree
Have you tried to factor the LHS yet?

- Jul 25th 2011, 03:23 PMRK29Re: Inequality with higher degree
I tried doing X(X^3 -1) <0 then x=0 x^3-1=0 didn't work when i checked using solution

- Jul 25th 2011, 03:28 PMpickslidesRe: Inequality with higher degree
Maybe you can factor $\displaystyle \displaystyle x^3-1$ further?

Hint: $\displaystyle \displaystyle a^3-b^3 = (a-b)(a^2-ab+b^2)$

Sketching the graph of $\displaystyle f(x) = x^4-x$ will help after you have found the zeros. - Jul 25th 2011, 03:40 PMPlatoRe: Inequality with higher degree
- Jul 25th 2011, 03:48 PMArchie MeadeRe: Inequality with higher degree
- Jul 25th 2011, 03:58 PMAlso sprach ZarathustraRe: Inequality with higher degree
- Jul 25th 2011, 04:54 PMskeeterRe: Inequality with higher degree
$\displaystyle x^4 - x < 0$

$\displaystyle x(x^3 - 1) < 0$

$\displaystyle x(x-1)(x^2 + x + 1) < 0$

note that the factor $\displaystyle (x^2 + x + 1) > 0$ for all x (how can you check that the factor is always positive?)

for the other two factors that can equal zero ... critical values are x = 0 and x = 1

these two critical values divide the x-value number line into three regions ...

x < 0 , 0 < x < 1 , and x > 0

test a single x-value from each of the above intervals into the original inequality to see if this value makes the inequality true or false ...

if true, then all values of x in that interval make the inequality true ... if false, then otherwise. - Jul 25th 2011, 05:41 PMtheloserRe: Inequality with higher degree
$\displaystyle x^4-x<0$

$\displaystyle x^4-x=0$

This works as well just remember to put the strict inequity back in the answer.