Function composition fraction question

Big_Blue824

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Hello!

So I'm doing the thing, and its making sense, but in g(f(x)), theres one bit where I'm not following the book. In the second line of work the book shows, right where my finger is pointing, at the numerator fraction's numerator, the 2 has disappeared from the
2-(x+1)
I've no idea why.
Can someone explain why it's gone?
 

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Hello!

So I'm doing the thing, and its making sense, but in g(f(x)), theres one bit where I'm not following the book. In the second line of work the book shows, right where my finger is pointing, at the numerator fraction's numerator, the 2 has disappeared from the
2-(x+1)
I've no idea why.
Can someone explain why it's gone?
Disappearing 2 is not the only difference. Can you simplify the first expression and see whether you get the second?
 
Direct Answer: 2 - (x+1) = 2 - x - 1 = 2 - 1 - x = (2-1) - x = 1 - x = -x + 1

Additional Response. TERRIBLE NOTATION.

(Ranting and raving deleted.) :)
 
Last edited:
Additional Response. TERRIBLE NOTATION.

[math]\dfrac{1}{x+1} + 3\cdot\dfrac{x+1}{x+1}[/math] -- No. Absolutely not. Do not allow this. Get a different book.

[math]\left[\dfrac{1}{x+1} + 3\right]\cdot\dfrac{x+1}{x+1}[/math] -- World of Difference! The numerator has the same problem. Just awful.
No, that's not what they mean! They are multiplying only the 3 by [MATH]\frac{x+1}{x+1}[/MATH] (and similarly in the numerator, multiplying only the 1), to make a common denominator.
 
No, that's not what they mean! They are multiplying only the 3 by [MATH]\frac{x+1}{x+1}[/MATH] (and similarly in the numerator, multiplying only the 1), to make a common denominator.
Thank you. This restores some faith in publishers.
 
2-1 = 1
You need to understand that from one line to the next line something will change so why are you surprised that the 2 disappeared?
 
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