Functions

[Loki123]

Is this correct?

 

[JeffM]

Your cancellation is wrong, but somehow you got the right answer for g(x).

[math]g(x) = \dfrac{f(x) - 1}{f(x) + 1} \text { and } f(x) = \dfrac{x- 2017}{x + 2017} \implies\\ g(x) = \left ( \dfrac{x - 2017}{x + 2017} - 1 \right) \div \left ( \dfrac{x - 2017}{x + 2017} + 1 \right) \implies \\ g(x) = \dfrac{x - 2017 - (x + 2017)}{x + 2017} \div \dfrac{x - 2017 + x + 2017}{x + 2017} \implies \\ g(x) = \dfrac{\cancel 2 * (-2017)}{\cancel {x + 2017}} \div \dfrac{\cancel 2 x}{\cancel {x + 2017}} = \dfrac{-2017}{x}.[/math]
Same issue when calculating f(g(x)), but once again you come out right.

 

[Loki123]

Your cancellation is wrong, but somehow you got the right answer for g(x).

[math]g(x) = \dfrac{f(x) - 1}{f(x) + 1} \text { and } f(x) = \dfrac{x- 2017}{x + 2017} \implies\\ g(x) = \left ( \dfrac{x - 2017}{x + 2017} - 1 \right) \div \left ( \dfrac{x - 2017}{x + 2017} + 1 \right) \implies \\ g(x) = \dfrac{x - 2017 - (x + 2017)}{x + 2017} \div \dfrac{x - 2017 + x + 2017}{x + 2017} \implies \\ g(x) = \dfrac{\cancel 2 * (-2017)}{\cancel {x + 2017}} \div \dfrac{\cancel 2 x}{\cancel {x + 2017}} = \dfrac{-2017}{x}.[/math]
Same issue when calculating f(g(x)), but once again you come out right.

thanks!

 

[Steven G]

The first function seems to me to be called F(x) while in g(x) it is called f(x).
You do realize that F(x) and f(x) do not need to be the same function.
Good work otherwise.