How do you find the domain of #h(x)= sqrt( 4-x^2)/( x-3)#?

1 Answer
George C.
Jun 11, 2015

The domain is restricted by the numerator to #[-2, 2]# and by the denominator to #(-oo,3) uu (3,oo)#, which is no additional restriction. So the domain of #h(x)# is #[-2, 2]#

Explanation:

In order for the numerator of #h(x)# to be defined, #4 - x^2 >= 0#.

Adding #x^2# to both sides we get #x^2 <= 4 = 2^2#

hence #-2 <= x <= 2#

In order for the denominator to be defined and non-zero, we just need #x != 3#

This condition only affects values of #x# outside the #[-2, 2]# range we already know the domain is restricted to.

So #h(x)# is well defined for all #x in [-2, 2]#.

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