So I have a functions question.. again.
$f(x) \rightarrow \frac{2}{x-1}$ and $x\neq{}-\frac{b}{a}$
Show that $f(p) + f(-p) = 2f(p^2)$.
My Work:
$\frac{2}{x-1}+\frac{2}{-x-1}$ which is supposedly equal to $\frac{-2-2x}{1-x}+\frac{2x-2}{1-x}$
This equals $\frac{-4}{1-x}$, but if you do the math to find $2f(p^2)$, you get (or at least I got) $\frac{-4x+4}{x^2-x^3+x-1}$.
I'm sure I did something wrong, but I've no clue what. Anyone care to assist?
$\endgroup$ 12 Answers
$\begingroup$we have $$f(x)=\frac{2}{x-1}$$ and we have to calculate: $$f(p)+f(-p)=\frac{2}{p-1}+\frac{2}{-p-1}=\frac{2}{p-1}+\frac{-2}{p+1}=\frac{4}{p^2-1}=2f(p^2)$$
$\endgroup$ $\begingroup$$\frac{2}{x-1}+\frac{2}{-x-1}$ is definitely not equal to $\frac{-4}{1-x}$.
$\frac{2}{x-1}+\frac{2}{-x-1} = \frac{2}{x-1} - \frac{2}{x+1} = \frac{2(x+1) - 2(x-1)}{(x-1)(x+1)} = \frac{4}{x^2-1} = 2 \frac{2}{x^2-1} = 2f(x^2)$
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