1. ## Domain again

Find the domain of this

$\displaystyle (-infinity,0] U [3,infinity)$ becuase its to the 4th root we have a restirction unlike an odd root which is all real number
why am I wrong ?

2. ## Re: Domain again

let's look at it this way:

let $\displaystyle g(x) = \sqrt[4]{x}$

let $\displaystyle h(x) = x^2 - 9$

then $\displaystyle f = g \circ h$

now h is clearly defined for all x, no problem there.

but g is only defined for x ≥ 0.

that means that f is only defined for h(x) ≥ 0.

so we need to look at what the RANGE of h is, in particular, for which x is h(x) ≥ 0?

this is precisely when:

$\displaystyle x^2 \geq 9$

that is:

$\displaystyle x \in (-\infty, -3] \cup [3,\infty)$.

to see why your answer is wrong, let's pick something in it, and see what happens:

let's use x = -1, which is in the interval (-∞,0].

now x2 - 9 = (-1)2 - 9 = 1 - 9 = -8. how are we going to take a 4th root of that?

EDIT: darn, i left out an x!

it should be:

x2 - 9x (is my memory, or my eyesight, going?)

this is ≥ 0 when either:

x ≥ 0 and x - 9 ≥ 0, that is x ≥ 9...in this case the "9" controls (since its bigger and we have to have both), so this is the interval [x,∞).

or:

x < 0 and x - 9 < 0, that is x < 9...in this case the "0" controls (since it is smaller), so this is the interval (-∞,0].

therefore, the correct answer is as Soroban said:

(-∞,0] U [9,∞).

3. ## Re: Domain again

Hello, M670!

$\displaystyle \text{Find the domain: }\;f(x) \;=\;\sqrt[4]{x^2-9x}$

We see that $\displaystyle x^2 - 9x$ must not be negative.

When is $\displaystyle x^2-9x$ greater than or equal to zero?

Consider the parabola $\displaystyle y \:=\:x^2-9x$.
When is it above the $\displaystyle x$-axis?
Code:
        |
* |             *
|
*|            *
----*-----------*----
0| .       . 9
|    . .
|
Domain: .$\displaystyle (\text{-}\infty,\:0\:\!] \cup [\:\!9,\:\infty)$