find domain of: 1/ [(the 4th root of)x^2 -9x]

Question

precal · Answer

set the denominator equal to zero and solve it.  Recall division can not have a zero, so the domain is all of the real numbers except those that make it zero.

anonymous · Answer

okay..does the 4th root change anything? thats what is confusing me

anonymous · Answer

$$x \neq 0$$$$x \neq 9$$

jhonyy9 · Answer

no,the 4th root not change nothing

anonymous · Answer

ok

precal · Answer

I meant to say the denominator can not be equal to zero and division by zero is undefined.

anonymous · Answer

so i would set: $$\sqrt[4]{x^2 -9x}$$ =0?

anonymous · Answer

and solve for that

precal · Answer

I meant to say the denominator can not be equal to zero and division by zero is undefined.

precal · Answer

Yes take both sides to the 4th power and that would get rid of the (1/4) power.

anonymous · Answer

$$x^2-9x \neq0$$

anonymous · Answer

so x^2 - 9x cant equal zero...so it equals zero at....$$\sqrt{9x}$$ ? haha

precal · Answer

I meant to say the denominator can not be equal to zero and division by zero is undefined.

precal · Answer

factor $$x^2-9x$$

anonymous · Answer

ok

precal · Answer

I meant to say the denominator can not be equal to zero and division by zero is undefined.

precal · Answer

x(x-9) therefore x can not be zero and x can not be 9 but x can be any other real number.

anonymous · Answer

i see...so i didnt need to solve for x, just factor? to get x(x-9)

anonymous · Answer

$$x(x-9)\neq0 \rightarrow x \neq 9,  x \neq 0$$

precal · Answer

I meant to say the denominator can not be equal to zero and division by zero is undefined.

precal · Answer

You are solving for zero.

myininaya · Answer

the domain is 
$$x^2-9x>0$$

anonymous · Answer

okay...i think i can take it from here..hopefully haha, thanks guys

precal · Answer

I meant to say the denominator can not be equal to zero and division by zero is undefined.

precal · Answer

what about x=9, x can not be 9, it would create a zero in the denominator.  You can not have a zero in the denominator.

anonymous · Answer

The domain is $x^2-9x \ge 0$.

myininaya · Answer

right thats why we have > and not >=

myininaya · Answer

anwar we don't want the bottom to be zero

myininaya · Answer

$$x^2-9x>0$$

anonymous · Answer

$$x <0, x >9$$

anonymous · Answer

how would i express x cant equal 0 and 9 in interval notation...im having trouble with that

anonymous · Answer

Oh it's in the denominator, I didn't see that :)

anonymous · Answer

you express it as 
$$(-\infty,0)\cup (9,\infty)$$

anonymous · Answer

hello all!

anonymous · Answer

ahh...i see where i made my mistake entering it, thank you!

anonymous · Answer

And hello to @myininaya and @satellite73 :)

myininaya · Answer

x(x-9)>0
---------------------------
x(x-9)=0 when x=0 or x=9

f(-1)=+
f(1)=-
f(10)=+

so the intervals for which the function exist are 
$$(-\infty, 0) U (9,\infty)$$
oops what satellite has lol

myininaya · Answer

hey anwar! :)

anonymous · Answer

lol
@myininany 
\cup

myininaya · Answer

shhh

anonymous · Answer

@satellite73: You should include the real numbers between $0$ and $9$.

myininaya · Answer

no he shouldn't

anonymous · Answer

well not really since you cannot take the fourth root of a negative number

anonymous · Answer

@Anwara
NO

anonymous · Answer

Oh no, these are the negative values, my bad!

myininaya · Answer

lol

anonymous · Answer

ok ;)

anonymous · Answer

haha..thanks guys i appreciate the help!

anonymous · Answer

you're welcome ;)