Help w/ function: If f(x) = x - 2x^2, what then is f(1/(2x+1)) ?

kaanj

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Jun 28, 2016
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The exercise: If f(x) = x - 2x^2

What is then f(1/(2x+1))

Answer should be (2x-1)/(2x+1)^2

I think it is very difficult. Probably, there is a rule which I don't know about. I do something like this:

1/(2x+1) - 2(1/(2x+1))(1/(2x+1))

1 / (2x+1) - 2(1 / (2x^2+4x+1))

How to continue from here? And should I have another approach?

I am very thankful for help or guidance on exact what subjects I should study to solve these "simple" algebraic things. Peace!
 
The exercise: If f(x) = x - 2x^2

What is then f(1/(2x+1))

Answer should be (2x-1)/(2x+1)^2

I think it is very difficult. Probably, there is a rule which I don't know about. I do something like this:

1/(2x+1) - 2(1/(2x+1))(1/(2x+1))

1 / (2x+1) - 2(1 / (2x^2+4x+1))

How to continue from here?
Convert the fractions to a common denominator (here) and combine (here).

To get you started, first note that (2x + 1)^2 is not equal to 2x^2 + 4x + 1 (here). Convert to a common denominator:

. . . . .\(\displaystyle \dfrac{1}{2x\, +\, 1}\, -\, \dfrac{2}{(2x\, +\, 1)^2}\)

. . . . .\(\displaystyle \left(\dfrac{1}{2x\, +\, 1}\right)\, \left(\dfrac{2x\, +\, 1}{2x\, +\, 1}\right)\, -\, \dfrac{2}{(2x\, +\, 1)^2}\)

...and so on. ;)
 
Convert the fractions to a common denominator (here) and combine (here).

To get you started, first note that (2x + 1)^2 is not equal to 2x^2 + 4x + 1 (here). Convert to a common denominator:

. . . . .\(\displaystyle \dfrac{1}{2x\, +\, 1}\, -\, \dfrac{2}{(2x\, +\, 1)^2}\)

. . . . .\(\displaystyle \left(\dfrac{1}{2x\, +\, 1}\right)\, \left(\dfrac{2x\, +\, 1}{2x\, +\, 1}\right)\, -\, \dfrac{2}{(2x\, +\, 1)^2}\)

...and so on. ;)

Of course! Making a fraction of everything makes it so much easier! Thank you so much!
 
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