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!
$\displaystyle \text{Find the domain: }\;f(x) \;=\;\sqrt[4]{x^2-9x}$
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?
Domain: .$\displaystyle (\text{-}\infty,\:0\:\!] \cup [\:\!9,\:\infty)$Code:| * | * | *| * ----*-----------*---- 0| . . 9 | . . |