1. ## Inequality

x^3+5x^2+2x-8/x^3-7x^2+7x+15<=0 <= means less then or equal to

2. ## Re: Inequality

Originally Posted by iFuuZe

x^3+5x^2+2x-8/x^3-7x^2+7x+15<=0 <= means less then or equal to
You need to use brackets where they're needed so we can understand what the inequation actually is.

Is it \displaystyle \displaystyle \begin{align*} \frac{x^3 + 5x^2 + 2x - 8}{x^3 - 7x^2 + 7x + 15} \leq 0 \end{align*}?

3. ## Re: Inequality

Originally Posted by iFuuZe

x^3+5x^2+2x-8/x^3-7x^2+7x+15<=0 <= means less then or equal to
Inequality is equivalent to the :

$\displaystyle \frac{(x-1)(x+4)(x+2)}{(x-5)(x-3)(x+1)} \leq 0$

4. ## Re: Inequality

yep so you factorice the top and bottom now what we do procede....

5. ## Re: Inequality

Originally Posted by iFuuZe
yep so you factorice the top and bottom now what we do procede....
A function will only ever stop being negative at the x intercepts, so evaluate where they are, then test some points between each of them to determine which of the regions give negative values for the function

7. ## Re: Inequality

Originally Posted by Prove It
A function will only ever stop being negative at the x intercepts...
or at a discontinuity.

8. ## Re: Inequality

Originally Posted by a tutor
or at a discontinuity.
Let's assume we've already established that \displaystyle \displaystyle \begin{align*} x \neq -1, 3, 5 \end{align*} :P

9. ## Re: Inequality

Originally Posted by Prove It
Let's assume we've already established that \displaystyle \displaystyle \begin{align*} x \neq -1, 3, 5 \end{align*} :P
Of course.

I'm just pointing out that you can get a sign change at a discontinuity too.

10. ## Re: Inequality

guys i dont understand, its better if i visualise a worked example if any can do that it would fantastic

11. ## Re: Inequality

Sign changes occur whenever the sign of one factor changes.

This will happen at x= -4,-2,-1,1,3 and 5.

So check x<-4. Here all factors <0 so the expression is positive.

x<-4 positive
-4<x<-2 negative
-2<x<-1 positive
-1<x<1 negative
1<x<3 positive
3<x<5 negative
x>5 positive