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

Answer 1

Nathan Axel

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

The role is

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

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

Consequently,

#4-x^2>0#

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

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

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

Consequently,

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)#

plot{1/sqrt(4-x^2) [-10.625, 10.375, -1.96, 8.04]}

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Answer 2

Hazel Anglin

The domain of the function (F(x) = \frac{1}{\sqrt{4 - x^2}}) is ([-2, 2]), and the range is ([0, +\infty)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.