Solve the inequality f(x)>-1. f(x)=(x-1)(x+1)/(x+2)(x-2)
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We have to solve f(x) > -1 for f(x)=(x-1)(x+1)/(x+2)(x-2)
f(x) > -1
=> (x-1)(x+1)/(x+2)(x-2) > -1
=> (x-1)(x+1) > -1*(x+2)(x-2)
=> x^2 - 1 > -1* (x^2 - 4)
=> x^2 - 1 > -x^2 + 4
=> 2x^2 > 5
=> x^2 > 5/2
=> x > sqrt 5/2 or x < -sqrt 5/2
Therefore x lies in (-inf , -(sqrt 5)/2) and ((sqrt 5)/2 , + inf)
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For the beginning, we'll re-write the numerator and denominator of the function as differnces of squares:
f(x) = (x^2 - 1)/(x^2 - 4)
Now, we'll substitute f(x) by it's expression in the given inequality:
(x^2 - 1)/(x^2 - 4) > -1
We'll add 1 both sides, and we'll multiply it by the denominator (x^2-4).
(x^2 - 1) / (x^2 - 4) +1>0
(x^2 - 1 + x^2 - 4)/(x^2-4)>0
We'll combine like terms:
(2*x^2-5)/(x^2-4)>0
We'll consider the numerator and denominator as 2 functions:
The numerator: f1(x)=2*x^2-5
The denominator f2(x)=x^2-4
We'll check the monotony of the numerator. In order to do so, first we'll find out the roots of the equation f1(x)=0
2*x^2-5=0 => 2*x^2=5 => x^2=5/2
x1= sqrt (5/2) and x2=-sqrt (5/2)
f1(x) is negative over the range (-sqrt5/2 ; sqrt5/2) and it is positive over the ranges (-inf., -sqrt5/2)U(sqrt5/2;+inf.)
We'll discuss the monotony of the denominator f2(x)=x^2-4
f2(x)= (x-2)(x+2)
(x-2)(x+2)=0
x1=2 and x2=-2
f2(x) is negative over the range (-2 ; 2) and it is positive over the ranges (-inf., -2)U(2;+inf.)
f2(x)>0 for x belongs to (-inf,-2)U(2,inf)
f2(x)<0 for x belongs to (-2,2)
f(x) > -1 if x belongs to the ranges (-inf,-2) U (-sqrt(5/2),sqrt(5/2)) U (2,inf).
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