What is the domain for (x^2-8x)^(1/4)?

Hello

Raising to 1/4 means take the square root of the square root.

You cannot take the square root of a negative number, it is not defined.

So, x² - 8x must be ≥ 0.

(x)(x - 8) ≥ 0

Make a short table

Between -inf and 0, x is negative. Over 0, it is positive
Between -inf and 8, x - 8 is negative. Over 8, it is positive

So, up to 0 both factors are negative ==> the product is positive
Between 0 and 8, x is positive, but x-8 negative ==> the product is negative
Over 8, both factors are positive ==> the product is positive

The function is defined when x ≤ 0 and when x ≥ 8.

Domain in interval notation : (-inf , 0] U [8, + inf)

Hope it helps

Bye !

yes-- all real numbers will work