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Functions Question.... Miscalculation??

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by DanaJ » Thu May 14, 2009 9:25 pm
When such a problem pops up, just remember to start from the beginning and regard f(3x + 1) as any other x:

g(f(3x+1)) = [f(3x + 1)]^2

Now, f(3x + 1) = 3(3x + 1) - 1 = 9x + 2

Use the above to get:

g(f(3x+1)) = (9x + 2)^2 = 81x^2 + 36x + 4.
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by mkbigmoz » Thu May 14, 2009 10:18 pm
Dana, I went through the function in the opposite manner.


Shouldn't the function simply replace the x?


fx= 3x-1, so shouldn't f(3x+1) = {3(3x-1)+1} = 9x-2


then gx= (9x-2)^2 = 81x^2 - 36x + 4



Why did you put the function of 3x-1 as 3(3x+1)-1? Shouldn't it have been 3(3x-1)+1?
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by DanaJ » Thu May 14, 2009 10:42 pm
Nope, I'm sorry, but you are mistaken.

Let me explain why.

You get that f(x) = 3x - 1.
Now, you're required to calculate f(3x + 1). To make it easier to understand, let's say that 3x + 1 = y. You'll agree that f(3x + 1) = f(y).

Since f(x) = 3x - 1, then f(y) = 3y - 1. Here you replace y = 3x + 1 to get:

3y - 1 = 3(3x + 1) - 1 = 9x + 2.

Hope this helps...
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