How do you find the domain and range of #f(x) = 2 / (1 - x²)#?

Question

Abigail Allen · Accepted Answer

Domain is defined by inequalities:
#x!=+-1# or a union of three intervals
#(-oo,-1)#, #(-1,1)#, #(1,+oo)#

Range is a combination of two intervals
#-oo < f(x) < 0# and #2 <= f(x) < +oo# Domain of this function is determined by the fact that denominator might be equal to zero and the function would be undefined in this case. So, we are talking about the following restriction on domain: #1-x^2 != 0# or #x^2 != 1# or #x != +-1# In other words, the domain is a union of three intervals: #-oo < x <-1#; #-1 < x < 1#; #1 < x < +oo# or, in another notation, #(-oo,-1)#, #(-1,1)#, #(1,+oo)# To determine the range, it's most useful to graph this function. The easiest way is to start by drawing a graph of a function #y=1-x^2#, invert it into #y=1/(1-x^2)# and multiply the latter by a factor 2 to get #y=2/(1-x^2)#. #y=1-x^2# graph{(1-x^2) [-5, 5, 10, 10, 5]} #y=1/(1-x^2)# graph{1/(1-x^2) [-10, 10, 5, 5, 10]} #y=2/(1-x^2)# graph{2/(1-x^2) [-10, 10, 5, 5, 10]} As you see, on the left and on the right graph is located below the X-axis and goes to negative infinity, while in the middle it has a minimum of #2# at #x=0# and from that goes to positive infinity. Because of this, the range is shown as two intervals: #-oo < f(x) < 0# and #2 <= f(x) < +oo# or, using a different symbol, #(-oo, 0)# and #[2,+oo)# Notice, that value #y=2# is included into the second interval.

Isaiah Anthony · Answer

undefined To find the domain and range of $ f(x) = \frac{2}{1 - x^2} $:

1. Domain: The function is defined for all real numbers except where the denominator becomes zero, which would result in division by zero. So, the domain is all real numbers except where $ 1 - x^2 = 0 $.
   
   $ 1 - x^2 = 0 $ when $ x^2 = 1 $.
   
   So, $ x = \pm 1 $.
   
   The domain of the function is all real numbers except $ x = \pm 1 $.

2. Range: To find the range, consider the behavior of the function as $ x $ approaches positive and negative infinity. As $ x $ approaches positive or negative infinity, $ 1 - x^2 $ approaches positive infinity.

Therefore, as $ x $ approaches positive or negative infinity, $ f(x) = \frac{2}{1 - x^2} $ approaches zero.

The range of the function is all real numbers except zero.