How do you find the domain of #f(x)=sqrt((x^2-16))#?

1 Answer
Apr 26, 2017

Answer:

#x >=4 and x<=-4 #

Explanation:

Domain for a function means the value/s of #x# for which the function is valid. In this case, the function is invalid if #(x^2-16)# is #<# 0 as there are no real values of the function. So, you find the values of #x# which would make #(x^2-16)# greater or equal to 0 as the function would be valid.
So,

# (x^2-16) >= 0 #
# x^2 >= 16 #
# x^2 >= 4^2 #
# |x| >= 4 #
Either,

#x >= 4 #

Or,

#x <= -4 #

So for the values of #x >=4 and x<=-4 #, #(x^2-16)# will be greater or equal to zero which will be give some value under square root making the function valid.
The remaining values # -4 < x < 4#, would give you a negative term for #(x^2-16)# which would not be valid under square root making the function invalid.

SO the domain is #x >=4 and x<=-4 #