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f(x)=x^2-f and g(x)= 2x+1 Find f(g(x)) help!
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\[f(x)=x^2-4 \]
\[g(x)=2x+1\]
Well ill try this: \[f(g(x))=f(2x+1)=(2x+1)^2-4=4x^2+4x+1-4=4x^2+4x-3\]
Woah... What?
I think I understand where you got the f(2x+1)? But I'm not sure...
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Basically when you say \(f(g(x))\), your referring to the operation of wherever x was in f(x), your putting \(g(x)\) instead!
So you would have \[f(g(x))=f(2x+1)=(2x+1)^2-4\] Because before, \(f(x)=x^2-4\) But now im putting g(x) where x was.
Okay, I see that. Now what for the next problem it wants me to find 2f(3)+3g(-2) I don't get it.
OOOOH Ouch that's okay. So: \[\eqalign{ &f(x)=x^2-4 \\ &g(x)=2x+1 \\ }\] Still right?
Yes,
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Would I just put 3 in for f(x) then it would make: \[2=3^2-4\]
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