http://webwork.mathstat.concordia.ca...7812eb1c11.png
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 ?
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http://webwork.mathstat.concordia.ca...7812eb1c11.png
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 ?
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 x^{2}  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:
x^{2}  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,∞).
Hello, M670!
Quote:
$\displaystyle \text{Find the domain: }\;f(x) \;=\;\sqrt[4]{x^29x}$
We see that $\displaystyle x^2  9x$ must not be negative.
When is $\displaystyle x^29x$ greater than or equal to zero?
Consider the parabola $\displaystyle y \:=\:x^29x$.
When is it above the $\displaystyle x$axis?
Domain: .$\displaystyle (\text{}\infty,\:0\:\!] \cup [\:\!9,\:\infty)$Code:
*  *

* *
**
0 . . 9
 . .
