Solve the inequality f(x)>-1. f(x)=(x-1)(x+1)/(x+2)(x-2)

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).