Question

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

Answer 1

`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]`