Question

What is the domain of the function: `f(x) =sqrt(x^2(x-3)(x-4))`?

Answer 1

`D_(f(x)) = (-oo,3]uu[4,+oo)`

Explanation:

Given
`color(white)("XXX")f(x)=sqrt(x^2(x-3)(x-4))`

To find the domain we need to determine which values of `x` are not valid.

Since the `sqrt("negative value")` is undefined (for Real numbers)

`x^2(x-3)(x-4) >= 0`

`x^2 >= 0` for all `x in RR`
`(x-3) > 0` for all `x>3, in RR`
`(x-4) > 0` for all `x>4, in RR`

The only combination for which
`color(white)("XXX")x^2(x-3)(x-4) < 0`
is when `(x-3) > 0` and `(x-4) < 0`

That is the only non-valid values for (Real) `x` occur when
`color(white)("XXX")x > 3` and `x < 4`

Answer 2

`(-oo, 3] uu [4, oo)`

Explanation:

The domain is where the radicand (the expression under the square root sign) is non-negative.

We know that `x^2 >= 0` for all `x in RR`.

So in order that `x^2(x-3)(x-4) >= 0`, we must either have `x^2 = 0` or `(x-3)(x-4) >= 0`.

When `x<=3`, both `(x-3) <= 0` and `(x-4)<=0`, so `(x-3)(x-4) >= 0`

When `3 < x < 4`, `(x-3) > 0` and `(x-4) < 0`, so `(x-3)(x-4) < 0`.

When `x >= 4`, both `(x-3)>=0` and `(x-4)>=0`, so `(x-3)(x-4) >= 0`.

So `x^2(x-3)(x-4)>=0` when `x in (-oo,3] uu [4, oo)`

Note that this domain already includes the point `x = 0`, so the `x^2 = 0` condition gives us no extra points for the domain.