Inequalities(Double equations)

Question

(x+1)(x-2)

---------------- ≥0

x^2+7x-8

Please explain this question.

Part of the answer is -1≤x≤1. But how to get the range of it? I can't figure it out. Thank you.

Answer 1

We are looking for where the rational expression is positive. The places where it could potentially change sign from positive to negative are when it equals zero and when it hits an asymptote.

It will equal zero when the numerator =0 (as long as the denominator isn't zero:

(x-1)(x-2)=0

x=1 x=2

It will not exist when the denominator =0

x²+7x-8=0

Factor

(x+8)(x-1)=0

x=1 and x=-8

Since both the numerator and the denominator =0 at x=1, this will be a hole not an asymptote. We have a zero at x=2 and an asymptote at x=-8

We now want to test places above, below and between these points to see if they are positive

x<-8

(x-1)(x-2)/(x+8)(x-1)

plug in -9 (any number less then -8 will do)

(-10)(-11)/(-1)(-10)

This will be positive since the negatives cancel

-8<x<2 (but not 1 b/c it doesn't exist)

(x-1)(x-2)/(x+8)(x-1)

plug in x=0

(-1)(-2)/(8)(-1)

This will be negative because the numerator will be positive and the denominator negative

x>2

(x-1)(x-2)/(x+8)(x-1)

plug in x=3 (or whatever)

(2)(1)/(11)(2)

All positive so positive.

This expression is true where x<-8 (not ≤ because -8 is an asymptote) and x≥2