What is the domain and range of #g(x)= (x^2 - 16)^(1/2)#?

1 Answer
Jun 9, 2018

Answer:

#x\in(-\infty, -4]\uu [4, \infty)#
#y\in[0,\infty]#

Explanation:

#g(x) = \sqrt{x^2-16}#

Anything underneath a square root must be greater than zero for the square root to be defined for real numbers.

#\therefore x^2-16>0 \Rightarrow x^2>16 \Rightarrow |x|>4#

#\therefore x\in(-\infty, -4]\uu [4, \infty)#

The range will be #[0,\infty)# since the expression inside the sqrt ranges from 0 to #\infty#

#\therefore y\in[0,\infty]#