Question

FUNCTION QUESTION

Answer 1

no more info about f?

Answer 2

Nope. That's all they give.

Answer 3

maybe like it's linear or something?

Answer 4

If I do that, then I get \[f(\frac{1}{1 - x}) + 2f(\frac{1 - x}{-x}) = \frac{1}{1 - x}\]

Answer 5

sorry i made a little mistake
wait a sec

Answer 6

look at x=2, x=1/2 and x=-1

Answer 7

f(2) + 2f(-1) = 2
f(1/2) + 2f(2) = 1/2
f(-1) + 2f(1/2) = -1

f(2) + 2f(-1) = 2
-2(f(-1) + 2f(1/2) = -1)

f(2) + 2f(-1) = 2
-2f(-1) - 4f(1/2) = 2

f(2) - 4f(1/2) = 4

f(1/2) - 4f(-1) = -4
f(2) - 4f(1/2) = 4

4f(1/2) - 16f(-1) = -16
f(2) - 4f(1/2) = 4

f(2) = -12 + 16f(-1)

f(2) = -12 + 16(2 - f(2))

f(2) = -12 + 32 - 16f(2)
f(2) = 20 - 16f(2)
17f(2) = 20
f(2) = 20/17
That's what I got :/

Answer 8

hmmm...not what i got

Answer 9

\[f(2)+2f(-1)=2\]
\[f(1/2)+2f(2)=1/2\]
\[f(-1)+2f(1/2)=-1\]


\[f(2)+2f(-1)=2\Rightarrow\frac{1}{2}f(2)+f(-1)=1\]
\[f(1/2)+2f(2)=1/2\Rightarrow 2f(1/2)+4f(2)=1\]

add the two equations on the right
\[\frac{1}{2}f(2)+f(-1)+2f(1/2)+4f(2)=1+1\]


\[\frac{1}{2}f(2)+-1+4f(2)=1+1\]

solve for \(f(2)\)

Answer 10

\[\frac{9}{2}f(2) = 3\]\[f(2) = \frac23\]Is that right?

Answer 11

yes...2/3

Answer 12

Thank you so much! I think I just made an arithmetic error somewhere :/. Again, thanks!

Answer 13

@Calcmathlete a sec is an hour now :D

let x=1/(1-x) , x=(x-1)/x u will have:

\[f(x)+2f(\frac{1}{1-x})=x\\f(\frac{1}{1-x})+2f(\frac{x-1}{x})=\frac{1}{1-x}\\f(\frac{x-1}{x})+2f(x)=\frac{x-1}{x}\]
three equation with three unknowns
\[f(x) \ \ \ and \ \ \ f(\frac{1}{1-x}) \ \ \ and \ \ \ f(\frac{x-1}{x}) \]
solving for f(x) gives

\[f(x)=\frac{1}{9}(\frac{4(x-1)}{x}+\frac{2}{x-1}+x) \]

Answer 14

lol. Thanks :D