banksy wrote:

Is x < 0 ?

(1) x^3 < x^2

(2) x^3 < x^4

Can someone explain how to find the ranges using the number line approach (+ve and -ve curves); I find these types so tough.

1.

x^3 < x^2

x^3 - x^2 < 0

x^2(x-1) < 0

Roots are; 0 and 1

Now; for a very small value of x; the above equation holds true

So; the ranges are x<0, 0<x<1, x>1

According to the rule; we should consider the -ve curves for "<" sign

so; x<0 and x>1. However, the correct ranges are x<0 or 0<x<1.

How come the positive curve 0<x<1 holds true for "<" sign and x>1 doesn't; How does the rule change for even exponents and odd exponents.

(2) x^3 < x^4

x^3 - x^4 <0

x^3(1-x) < 0

x^3(x-1) > 0 {Multiplying by -ve}

roots 0 and 1

for a big value of x, the above equation holds good

so; the ranges are; x>1, 0<x<1, x<0

Considering +ve curves because the sign here is ">"

x>1

x<0

This looks okay.

You should simplify x^3 < x^2 first. Reduce by x^2 (x^3 < x^2 means that x is not equal to zero so x^2>0) --> x<1, excluding one point x=0.

So you can ignore the even power expressions in the approach you talk (just look whether zero is inclusive or not). For example, solve (x+3)(x+1)(x-1)(x+5)^2<0:

So we have (x+3)(x+1)(x-1)<0 and x doesn't equal to -5 --> roots -3, -1 and 1. Trying extreme value for x tells that in 4th range (when x>1) the whole expression is positive so it's negative in 3rd and 1st ranges (+-+-): -1<x<1 or x<-3 excluding x=-5 (you can wirte it also as x<-5, or -5<x<-3, or -1<x<1).

If it were: (x+3)(x+1)(x-1)(x+5)^2<=0 then the answer would be: -1<=x<=1 or x<=-3 (so x=-5, as well as the other roots, is included as the whole expression can equal to zero)