anishasjkaul wrote:
Hi, Bunuel
The trick for inequalities written in the first post of this link, is it always true??... Will it work for all inequalities? Even third degree equation of X ( for example) with an inequality sign?
Yes, it is always true. But a more easy approach is as follows:
Lets have an inequality of degree n of the form
(x-1)(x-2)(-3-x)(x-4)<=0,
Now here, we see that coefficient of highest power of x is negative. My approach is as follows:
Step 1: Make coefficient of highest power positive by changing the sign of inequality
(x-1)(x-2)(x+3)(x-4)>=0
Step 2: Identify critical point.
Four critical points are there -3,1,2,4
Since you have already made coefficient of highest power of x positive in first step, the inequality if always be positive beyond the right most critical point i.e. beyond 4 it is always positive
Step3: Plot the regions.
Now the +ve and -ve regions will alternate
>4 it is positive
2<x<4 it is negative
1<x<2 it is positive
-3<x<1 it is negative
x<-3 it is positive.
In this way you can solve the inequality.
In case we have an even power of a term as shown below:
(x-1)(x-2)^2(x-3)(x-4)>=0
The regions at the critical point 2 will not alternate. We will again start from right most critical point:
x>4 is positive
3<x<4 is negative
2<x<3 is positive
1<x<2 is positivex<1 is negative
In case of an odd power the signs alternate normally.
Hope it makes sense!!!
Kudos if you like!!!