F(x)+f(1/x)=x, find f(x) in terms of x. Pic unrelated

f(x)+f(1/x)=x, find f(x) in terms of x. Pic unrelated

what about x=0?

x≠0

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

hmm

f(x) + f(1/x) = x (1)

let x --> 1/x

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

Setting (1) = (2)

x = 1/x

x^2 = 1

x = -1, +1

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

What the hell are you on?

are you retarded brainlet? such a function can only be well-defined at x= 1 and x=-1

...

lim = inf + 0 = inf
(x->inf)

[math] f(x) = x - f(\frac{1}{x}) [/math] and [math] f(\frac{1}{x}) = \frac{1}{x} - f(x) [/math]. Plugging one into the other yields
[eqn] f(x) = x - \frac{1}{x} + f(x) \iff x = \frac{1}{x} \iff |x| = 1 [/eqn]

Knowing this, we know that the domain of f is [math] \{ -1,1 \} [/math]. So now we can focus on these numbers.

[eqn] f(1) = 1 - f(1) \iff f(1) = \frac{1}{2} [/eqn]
[eqn] f(-1) = -1 - f(-1) \iff f(-1) = -\frac{1}{2} [/eqn]

Which, as I have proven, is the unique function determined by your constraints.