f(x) = 10-x and g(x) = 5x+6 , find the following: a) f(g(x)) b) g(f(x)) c) f(f(x))

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f(x) = 10-x and g(x) = 5x+6 , find the following: a) f(g(x)) b) g(f(x)) c) f(f(x))

Algebra
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Just giving me an example of a) Is fine too im just trying to figure out how to set it up and solve it ^^
f(g(x)) means to simply put the expression, g(x), into f(x) in place of all instances of the variable x. Think of it like a search-and-replace, where every x in the "outer function" (in this case, f(x)) is replaced by the inner function, in this case (5x + 6). So you get f(g(x)) = 10 - (5x + 6) = 4 - 5x Now can you try the other 2?
f(g(x)) just applies you put the function g(x) where ever there is an x in f(x). So, \[f(g(x)) = 10-(5x+6)\] do so similarly with others.

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@DebbieG Ok so you got 4 - 5x by 10 - 6 and 5x being by it self? so b) would be 5(10 - x) + 6 = 50x + 6 ?
Your distribution is wrong, but yet you get the point :)
yes*
\[g(f(x)) = 5(10-x)+6 \implies 50-5x+6\]
Yes, 5(10 - x) + 6, but 5(10 - x) + 6 != 50x + 6 I don't really understand what you mean by, "Ok so you got 4 - 5x by 10 - 6 and 5x being by it self?"
@DebbieG I was trying to see if i understood how you did it with my way of explaining how i took what you did. If that clears it up
@Astrophysics Ok so would that mean f(f(x)) 10 - (10 - x) = 10 - 10x?
\[f(f(x)) = 10 - (10-x)\]
@Astrophysics Thank you so much ^^ That was easier then i thought lol
Note that there is a "negative 1" there
|dw:1434883474751:dw|
Np :P
there is another notation for these problems a) f(g(x)) <--- the same as f o g b) g(f(x)) <--- the same as g o f c) f(f(x)) <--- the same as f o f Looks like more textbooks are starting to get rid of the letter circle letter notation
Ah yes, when I was learning this, we had to do it both ways
I think it may have confused people, because it was written as such f o g (x)
I had to read right to left when I had that in my textbook
mine was strictly the letter o letter notation
xD Yeah, it was tricky at first

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