SOLUTION: let a function f be defined by f(x) = x^2-2x-3/x^2+2x+3 i) determine the domain of f(x) ii) find the range of f(x)

If there are no denominators or even roots (such a square roots), when solved
for y, the domain would simply be all real numbers or 

(-∞,∞)

Yours has no even roots when solved for y. The domain is the set of all possible
values of x such that no denominator is equal to 0.  So we set your denominator
equal to 0 to see what, if any, numbers we must exclude from 
(-∞, ∞) 



That doesn't factor so we use the quadratic formula. 







That's going to give us a negative number -8 under the square root, so it has no
real zeros.  Therefore there are no real numbers to exclude from (-∞, ∞), and so
that is the domain,  

Domain: all real number or (-∞, ∞)

----------------------

To find the range we put y for f(x) and solve for x:











 

What's under the square root must be ≧ 0

 (2y+2)^2-4*(y-1)*(3y+3) ≧ 0

Also the denominator must not be 0, so y ≠ 1

So 1 is a critical value.

 (2y+2)^2-4*(y-1)*(3y+3) ≧ 0 

4y^2+8y+4-4(3y^2-3) ≧ 0

4y^2+8y+4-12y^2+12 ≧ 0

-8y^2 + 8y + 16 ≧ 0

Dividing through by -8 reverses the inequality:

y^2 - y - 2 ≦ 0

(y-2)(y+1) ≦ 0

Critical values 2,-1 and 1 from above.

2 and -1 only cause what's under the square root to be 0, but do not cause the
denominator to be 0.  So they are part of the range. But 1 is not part of the
range since it causes the denominator to be 0.

-------●-----○--●------
-3 -2 -1  0  1  2  3  4 
 
Test the region left of -1 with test value -2

(y-2)(y+1) ≦ 0
(-2-2)(-2+1) ≦ 0
(-4)(-1) ≦ 0
4 ≦ 0
That's false so we don't shade left of -1

Test the region between -1 and 1 with test value 0

(y-2)(y+1) ≦ 0
(0-2)(0+1) ≦ 0
(-2)(1) ≦ 0
-2 ≦ 0
That's true so we shade between -1 and 1:

-------●=====○--●------
-3 -2 -1  0  1  2  3  4 


Test the region between 1 and 2 with test value 1.5

(y-2)(y+1) ≦ 0
(1.5-2)(1.5+1) ≦ 0
(-0.5)(1) ≦ 0
-0.5 ≦ 0
That's true so we shade between 1 and 2:

-------●=====○==●------
-3 -2 -1  0  1  2  3  4 

Test the region right of 2 with test value 3

(y-2)(y+1) ≦ 0
(3-2)(3+1) ≦ 0
(1)(4) ≦ 0
4 ≦ 0
That's false so we don't shade right of 2.

The range is given by this graph:

-------●=====○==●------
-3 -2 -1  0  1  2  3  4   

In interval notation:

    [-1,1) ᑌ (1,2]

However there are cases where a graph crosses its horizontal asymptote.  
So even though the graph has a horizontal asymptote at y=1, we must check
to see if it crosses its horizontal asymptote.

Here is the graph.  Notice that all values of x are used so that's why its
domain is (-∞, ∞).  

Also notice that the graph is never higher than 2 nor lower than -1, but
reaches those values.  Also notice that even though there is a horizontal asymptote at y=1 (the green line), which shows that its range must include
y=1 after all.



We show that it does intersect its horizontal asymptote by setting y=1 and
solving for x:

















So since the graph crosses its horizontal asymptote, it contains
the point (-3/2,1) and therefore 1 is included in the range as well.

Therefore the range is [-1,2]

Edwin