could you please help me find the domain of f(x) = 1/ 4th root of (x^2-5x)?

Answer 1

x^2-5x>0
x(x-5)>0
x<=0 orx>5
all values of 5 and above or 0 or less
x>=5 or x<=0

Answer 2

You can only take the 1/4 root of non-negative numbers -- so the domain will be all the Real Numbers where

(x^2-5x) >= 0

This can be factored as x*(x-5)

This will be non-negative when x<= 0 oe when x >=5

It is not defined when 0 < x < 5

Answer 3

Hope I interpret yr q right. By domain you mean real values of x for which f(x) is real?

x^2-5x=x(x-5)

so for all x>5 your f(x) is +ve and has real fourth root.

Best luck