Let ƒ(x)=1x and g(x)=x2 − 6x. What two numbers are not in the domain of ƒ º g? Separate your answers with a comma.
1/x not 1x typo sorry
Do you mean? \[f(x)=\frac{ 1 }{ x }\]\[g(x)=x^2-6x\]
yes
please help!!
How do I figure it out? It's a timed question and I only have 2 min left
\[f(g(x))=f º g(x)\], substitute g(x) in the place of "x" for f(x). Please do it for me.
How? I'm so confused!
\[f(x)=\frac{ 1 }{ x }\]where g(x) is the input. \[f(g(x))=\frac{ 1 }{ (x^2-6x) }\]
so how do I find the answer?
Now you can simplify the denominator, can you do it for me please?
9?
how 9? The denominator is \[x^2-6x\]and you see both terms have "x" common, so you re-write it as\[x(x-6)\]. We know, we can't have division by zero, for which values of "x", the denominator becomes zero?
0.. I dont understand at all. I'm panicking because I'm running out of time and I need to get this answer correct in order to pass. Can u give me the answer and then explain to me?
x=0 or x=6 is the answer.
how come?
and thanks so much. it was correct
\[A*B=0\]is either of "A" or "B" is zero, you will get zero. \[0*B=0\]or\[A*0=0\]Now replace "A" with "x" and "B" with "x-6" and try to make each term zero individually.
Oh! Thanks so much!!
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