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Mathematics 11 Online
OpenStudy (mtalhahassan2):

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

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):

\[fg(x)=f(g(x))\]\[=f(x-1)\]\[=3(x-1)^2\]

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|

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

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 :)

OpenStudy (mtalhahassan2):

Still need help!!

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