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Question 163839: supposed f(x)=-4x+3 and g(x)=1/x+2. evaluated f(g(x)) and g(f(x)) for x. Note any restrictions. show your work.
THANK YOU!
Answer by midwood_trail(310)   (Show Source):
You can put this solution on YOUR website! Supposed f(x)=-4x+3 and g(x)=1/x+2. Evaluated f(g(x)) and g(f(x)) for x. Note any restrictions. Show your work.
f(g(x)) means to plug the value of g(x) in the function f(x) and simplify.
f(1/(x+2))....Everywhere you see an x for f(x), replace it with the value
of g(x), which is the fraction 1/(x+2).
This is what it looks like:
f(1/(x+2)) = -4(1/(x+2)) + 3
f(1/(x+2)) = -4/(x + 2) + 3
We can write 3 as 3/1 to make our fraction work easier.
We add -4/(x + 2) + 3/1 and get (3x + 2)/(x + 2) for f(g(x)).
The restriction is the domain. In other words, the restriction will be the value(s) we can safely replace x with in the final answer without producing division by zero.
For the fraction, (3x + 2)/(x + 2), x can be all real numbers
EXCEPT x CANNOT = -2. If you replace x with -2 in the above fraction, you will create division by zero, which does not exist or is UNDEFINED.
So, f(g(x)) = (3x + 2)/(x + 2) with restriction x = all real numbers EXCEPT that x cannot = -2.
Is this clear?
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To find or evaluate g(f(x)), we replace every x that you see for
g(x) with -4x + 3 and simplify.
g(f(x)) = g(-4x + 3)
Then:
g(-4x + 3) = 1/(-4x + 3) + 2
We simplify the denominator in the fraction 1/(-4x + 3) + 2 and we get:
1/(-4x + 5)
So, g(f(x)) = 1/(-4x + 5) with restriction or domain x = all real numbers
EXCEPT x cannot = 5/4 because the fraction 5/4 will
create division by zero in the fraction 1/(-4x + 5), which does not exist or is UNDEFINED.
Did you follow?
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