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#y=3/(2-x^2)#
We know that, in a rational function, at the #x#-values where the denominator becomes #0# the function becomes undefined. At those locations, we have vertical asymptotes.
#2-x^2=0, :. x=+-sqrt2#
Therefore, the function's domain will be in three pieces:
#(-oo < x < -sqrt2) and (-sqrt2 < x < sqrt2) and (sqrt2 < x < oo)#
Because the degree of the denominator is higher than the degree of the numerator, the #x#-axis is the horizontal asymptote. And because the numerator is a constant, #y# can never be #0#. This means the range of the function is not one piece. Rather, it is in two or more pieces. We need to examine the function more closely.
For all #x#-values in the part of the domain #(-oo < x < -sqrt2)#, the function is always negative and its range is:
#0 > y > -oo#
For the #x#-values in the part of the domain #(sqrt2 < x < oo)#, the function is always negative and its range is:
#-oo < y < 0#
For #x#-values in the part of the domain #(-sqrt2 < x < sqrt2)#, the function is always positive. It means this piece of the function has to have both ends of it going to #oo#, i.e. at #-sqrt2 and sqrt2#.
Then, it must be #U#-shaped and therefore must have a minimum point. We know #y# can not be #0#. As such, the minimum point must be somewhere above #y=0#.
The function becomes minimum when its denominator reaches its maximum value. The maximum value of #2-x^2# is when #x^2=0# which means #x=0 and y=3/2#. Therefore, the range of the function is:
#0 > y > -oo and 3/2 <= y < oo#
The following is the graph of the function where the domain and range identified above become obvious:
