Need help!!
In each case, function f and g are defined for xER. For each pair of functions, determine the expression and the domain of f(g(x)) and g(f(x)). Graph each result.
a) f(x)= 3 x^2, g(x)= x-1
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OpenStudy (anonymous):
This problem requires you to use a method called composition of functions.
So in the case of f(g(x)), you are putting g(x) into f(x). So that everywhere you see an x in f(x), you replace it with the function g(x).
So, f(g(x)) is 3(x-1)^2
Hopefully that gets you started
OpenStudy (mtalhahassan2):
ok so for the f(g(x)) = x-1
x-1 =3x^2+x-1
OpenStudy (mtalhahassan2):
so it be like this
OpenStudy (mtalhahassan2):
@Tuggaro
OpenStudy (anonymous):
\[fg(x)=f(g(x))\]\[=f(x-1)\]\[=3(x-1)^2\]
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OpenStudy (mtalhahassan2):
ok
OpenStudy (mtalhahassan2):
why u write 3
OpenStudy (anonymous):
bcoz u say that \[f(x)=3x^2\]
OpenStudy (mtalhahassan2):
but then where is x square go
OpenStudy (anonymous):
|dw:1424070102738:dw|
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OpenStudy (mtalhahassan2):
oh ok
OpenStudy (mtalhahassan2):
so know what I have to do
OpenStudy (anonymous):
\[gf(x)=g(f(x))\]can u do this?
OpenStudy (mtalhahassan2):
yeah sure
OpenStudy (mtalhahassan2):
g(f(x))=3x^2
= 3x^2+x-1
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OpenStudy (anonymous):
it suppose to be \[gf(x)=g(f(x))\]\[=g(3x^2)\]\[=3x^2-1\]
OpenStudy (mtalhahassan2):
nope i don't
OpenStudy (kohai):
Please close the question if you no longer need help :)