# Inequality with higher degree

• Jul 25th 2011, 03:15 PM
RK29
Inequality with higher degree
x^4-x<0

How do I solve it?
• Jul 25th 2011, 03:21 PM
pickslides
Re: Inequality with higher degree
Have you tried to factor the LHS yet?
• Jul 25th 2011, 03:23 PM
RK29
Re: 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 PM
pickslides
Re: 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 PM
Plato
Re: Inequality with higher degree
Quote:

Originally Posted by RK29
x^4-x<0 How do I solve it?

Can you see that $\displaystyle x\not= 0~?$
So can you solve $\displaystyle x^3-1<0~?$
If so solve it.
• Jul 25th 2011, 03:48 PM
Re: Inequality with higher degree
Quote:

Originally Posted by RK29
I tried doing X(X^3 -1) <0 then x=0 x^3-1=0 didn't work when i checked using solution

Hint: f(1)=0 and so (x-1) is a factor.
• Jul 25th 2011, 03:58 PM
Also sprach Zarathustra
Re: Inequality with higher degree
Quote:

Originally Posted by RK29
x^4-x<0

How do I solve it?

$\displaystyle x^4$ is non-negative number for all $\displaystyle x\in\mathbb{R}$

So when, $\displaystyle x^4<x$?

x must be positive, but not greater than $\displaystyle 1$. Can you see that?
• Jul 25th 2011, 04:54 PM
skeeter
Re: Inequality with higher degree
Quote:

Originally Posted by RK29
x^4-x<0

How do I solve it?

$\displaystyle x^4 - x < 0$

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

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

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 PM
theloser
Re: 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.