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Function f(1-x)
- Thread starter bobrossu
- Start date
D
Deleted member 4993
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(x2-2x-4)/(2-x) = [(x-1)2 - 5]/[1 - (x-1)]....... continue....Confused on what to do first.
Problem:
If f(1-x)=((x2-2x-4)/(2-x))
find f(1/x)
Work shown would be great!
(x2-2x-4)/(2-x) = [(x-1)2 - 5]/[1 - (x-1)]....... continue....
Why is it (x-1) and not (1-x) when substituting the x's in the function?
Also what happened to make the (x2-2x-4) into ((x-1)2 - 5) ?
**Edit**
Nevermind, I see what you did! Thanks!
Last edited:
Let's try it like this.Confused on what to do first.
Problem:
If f(1-x)=((x2-2x-4)/(2-x))
find f(1/x)
Work shown would be great!
\(\displaystyle u = 1 - x \implies x = 1 - u \implies x^2 = 1 - 2u + u^2.\)
\(\displaystyle \therefore f(1 - x) = f(u) = \dfrac{x^2 - 2x - 4}{2 - x} = \dfrac{(1 - 2u + u^2) - 2(1 - u) - 4}{2 - (1 - u)} \implies\)
\(\displaystyle f(u) = \dfrac{1 - 2u + u^2 - 2 + 2u - 4}{2 - 1 + u} = \dfrac{u^2 - 5}{1 + u}.\)
Let's check.
\(\displaystyle f(1 - x) = \dfrac{(1 - x)^2 - 5}{1 + (1 - x)} = \dfrac{1 - 2x + x^2 - 5}{2 - x} = \dfrac{x^2 - 2x - 4}{2 - x}.\)
It checks.
Therefore what is \(\displaystyle f \left ( \dfrac{1}{x} \right ).\)
Let's try it like this.
\(\displaystyle u = 1 - x \implies x = 1 - u \implies x^2 = 1 - 2u + u^2.\)
\(\displaystyle \therefore f(1 - x) = f(u) = \dfrac{x^2 - 2x - 4}{2 - x} = \dfrac{(1 - 2u + u^2) - 2(1 - u) - 4}{2 - (1 - u)} \implies\)
\(\displaystyle f(u) = \dfrac{1 - 2u + u^2 - 2 + 2u - 4}{2 - 1 + u} = \dfrac{u^2 - 5}{1 + u}.\)
Let's check.
\(\displaystyle f(1 - x) = \dfrac{(1 - x)^2 - 5}{1 + (1 - x)} = \dfrac{1 - 2x + x^2 - 5}{2 - x} = \dfrac{x^2 - 2x - 4}{2 - x}.\)
It checks.
Therefore what is \(\displaystyle f \left ( \dfrac{1}{x} \right ).\)
Is it f(1/x)=(1-5x2)/(x2+x) ?
Looks good to me.Is it f(1/x)=(1-5x2)/(x2+x) ?