Question

How do you find the domain for `f(x) = sqrt(4 - x^5)`?

Answer 1

Assuming we are restricted to Real numbers (i.e. `f(x) epsilon RR`)
Then the domain of `f(x)=sqrt(4-x^5)`
is all values of `x` for which the argument of the square root is non-negative.
That is
`x^5<= 4`
or
`x<= root(5)(4)`
So the domain of `f(x)` is `(-oo,root(5)(4)]`

Answer 2

Domain: All real numbers `<=` (4)`^1/5`

The domain of f(x) would be all real values of x for which 4-`x^5` is greater or equal to 0. The values of x which make 4-`x^5` negative cannot be accepted because in that case `sqrt(4-x^5)` would become imaginary.
Hence solving the inequality 4-`x^5` `>=` 0, it would be 4`>=` `x^5`. That is 4^`1/5` `>=` x

In the interval notation it would be ( -inf, 4^`1/5`]