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Just subtract for finding (f - g) and add for finding (f + g).
(f-g)(x) = f(x) - g(x)
\[\rm (f - g)(x) = (2x - 2)-(x + 4)\]\[\rm (f + g)(x) = (2x + 2) + (x + 4)\]

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Oh, then it gets a little difficult... you have to plug g(x) into f(x).
For example, if you got \(\rm f(x) = x^2\) and \(\rm g(x) = x+1\), then\[\rm f(g(x)) = f(x + 1) = (x + 1)^2\]
oh..ok :/
soo what would this be?
for f(g(x)) just replace all the x in 2x-2 with (x+4) and for g(f(x)) just replace all the x in x+4 with (2x-2)
\(x(f-g)\)=\(f(x)-g(x)\) now put the values of \(f(x)\) & \(g(x)\) & solve them.
yup.. thats how its done.
thanks!
welcome.
so would it then be 2x+8-2 8x-8?
how u got 8x-8?
2x times +4 and-2 times +4
noooo... (2x-2) +4 u have to just open the bracket.. if there is a positive or negative sign outside the bracket that means its addint or subtracting not multiplying. your expression will be like 2x-2 +4
so it would just be 2x-2 +4? i dont have to solve it?
(2x-2)4 if it was written liek this then u had to multiplied.. and yes that would be your answer just simplify it 2x+2 would be your answer for gf(x)
thanks!
welcome ..

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