What is the domain and range of #F(x) = 1/ sqrt(4 - x^2)#?

1 Answer
Jul 8, 2018

Answer:

The domain is #x in (-2,2)#. The range is #[1/2, +oo)#.

Explanation:

The function is

#f(x)=1/sqrt(4-x^2)#

What'under the #sqrt# sign must be #>=0# and we cannot divide by #0#

Therefore,

#4-x^2>0#

#=>#, #(2-x)(2+x)>0#

#=>#, #{(2-x>0),(2+x>0):}#

#=>#, #{(x<2),(x> -2):}#

Therefore,

The domain is #x in (-2,2)#

Also,

#lim_(x->2^-)f(x)=lim_(x->2^-)1/sqrt(4-x^2)=1/O^+=+oo#

#lim_(x->-2^+)f(x)=lim_(x->-2^+)1/sqrt(4-x^2)=1/O^+=+oo#

When #x=0#

#f(0)=1/sqrt(4-0)=1/2#

The range is #[1/2, +oo)#

graph{1/sqrt(4-x^2) [-9.625, 10.375, -1.96, 8.04]}