How do you find the domain of #y( x ) = \sqrt { 16- 8x }#?

1 Answer

The domain of the function is #x <=2# (That is, #x# is less than or equal to #2#).

Explanation:

The question requires us to find where the function is defined. We are aware that the function is a square root function. From our basic understanding of square roots, the value under the square root cannot be negative.

Following that logic, we introduce an inequality, that is, the function must be defined where

#16 -8x >= 0#

Solving this inequality, the result is #x <= 2#.