Please help. Will fan and medal. Question attached in comments
alright; replace x in f(x) with g(x)
brb
need to go eat
\[\sqrt{4 * 1/x-4}\] ?? and okay I'll be here whenever you're ready
Did you try graphing f o g?
I don't know how to ._.,
substitute g(x) in for f(x)
so you have f of g(x)...
Tat's what I was trying to do in that equation above...
(that's the function f (g(x)) )
Okay so how would it look the way you are saying? I've never seen it like that before I don't think so I'm not sure what would be substituted where exactly.
Write out f(x) and g(x). \Throw out the "x" from f(x), on both sides of the equal sign. Replace it with g(x) = 1/(x-4). This is your f(g(x)). Show your work, please.
\[f(x) = \sqrt{4x} \] \[g(x)=1/x-4\] \[f(1/x-4)=\sqrt{4*1/x-4}\] @mathmale
That x-4 must be enclosed in parentheses for max. clarity. Please do that. Combine the 4 and 1 to get 4.
Sooooo \[f(1/x-4) =\sqrt{4/(x-4)}\] ?
Is that right? :/
right
So what now?
You can graph that function or analyze it given your possible answers.
If x is negative you're going to get complex numbers right?
If x is 4 you get a function with a real number divided by zero
sqrt of -number will not be found in the domain of the graph amongst the reals
I'm kind of confused by those last three statements ._.,
So the Sqrt of 4/ x-4 will have a domain such that x-4 is greater than zero ... Does that make sense to you?
More than before but not fully
Ahh okay. So is 5 the answer?
Yeah. Cuz the square root of a negative number doesn't exist, and neither does the square root of 4/0 in the case of the domain, since you're only dealing with real numbers.
Graphing the function would confirm that 5 is the only number on the list in the domain.
Thank you.! Can you help me with some more?
Yeah sure.
Please post further questions separately, OK?
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