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

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Thread replies: 22
Thread images: 2

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f(x)+f(1/x)=x, find f(x) in terms of x. Pic unrelated
>>
what about x=0?
>>
>>8832486
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
>>
>>8832503
What the hell are you on?
>>
>>8832505
are you retarded brainlet? such a function can only be well-defined at x= 1 and x=-1
>>
File: IMG_20170416_214037.jpg (136KB, 1280x720px) Image search: [Google]
IMG_20170416_214037.jpg
136KB, 1280x720px
>>8832484
>>
lim = inf + 0 = inf
(x->inf)
>>
>>8832484
[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.
>>
>>8832484

Ill take the chubby one at the right most side.

She must have a good ass once you take those tight panties off right?
>>
>>8832556
>She
wew lad.
>>
>>8832558

W-what are you implying anon
>>
>>8832561
Look at their pants. Only the 2 girls on the left are women.
>>
>>8832563

nigga those are women
>>
>>8832484
>pic unrelated
thanks retard
>>
>>8832574
Well, I guess this is exactly why people say men are nothing but dreamers.

Dream on, space man.
>>
>>8832574
Look at the shoulders and body shape dawg
>>
>>8832484
I need that maxwell grill NOW
>>
>>8832582
>>8833027
Too late i already came before finishing the second on the left formulas
>>
>>8832484
Are solutions on the front
>>
You can do without any math. f(x) = -1/2 because that would reverse the 1/x shit.
Thread posts: 22
Thread images: 2


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