SOLUTION: If f(x) = 2x + 1 and g(x) = (x – 1)/2 , then f(g(x)) = So, I know what to do... Basic plug and chug. I also know that the answer is X, I just don't know how to get to the answer

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: If f(x) = 2x + 1 and g(x) = (x – 1)/2 , then f(g(x)) = So, I know what to do... Basic plug and chug. I also know that the answer is X, I just don't know how to get to the answer      Log On


   

Question 756543: If f(x) = 2x + 1 and g(x) = (x – 1)/2 , then f(g(x)) =
So, I know what to do... Basic plug and chug. I also know that the answer is X, I just don't know how to get to the answer.
I plug in g(x)...
f(x) = 2(x-1/2)+1, distribute... (2x-2+1)/2, combine like terms... (2x-1)/2.
Now that's where I get stuck. I don't know how to get to x, unless I did my math wrong. How do I get to x?
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
If f(x) = 2x + 1 and g(x) = (x – 1)/2 ,
then f(g(x)) = f[(x-1)/2] = 2[(x-1)/2]+1 = (x-1)+1 = x
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Cheers,
Stan H.
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Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
What all this says is whenever you see x in f%28x%29 then replace it with g%28x%29

f%28x%29=2x%2B1 ......if g%28x%29=%28x-1%29%2F2 then
f%28g%28x%29%29=2%28%28x-1%29%2F2%29+%2B1
f%28g%28x%29%29=cross%282%29%28%28x-1%29%2Fcross%282%29%29+%2B1
f%28g%28x%29%29=%28x-1%29+%2B1
f%28g%28x%29%29=x-1+%2B1
f%28g%28x%29%29=x