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Algebra Level 2

Given that $f(2^x)+xf(2^{-x})=1$ , find the value of $f(2)$ .

1 -2 0 -3 3 2 -1

5 solutions

Victor Loh
Jul 18, 2015

Sub $x=1$ , we get $f(2)+f\left(\frac{1}{2}\right)=1$ .

Sub $x=-1$ , we get $f\left(\frac{1}{2}\right)-f(2)=1$ .

First equation - Second equation:

$2f(2)=0 \implies f(2)=\boxed{0}.$

I used a similar approach:

$f(2^x) + x*f(2^{-x} )= 1 = f(2^{-x} ) +(-x)f(2^x)$

$\Rightarrow f(2^x) - (-x)f(2^x) + xf(2^{-x}) - f(2^{-x}) = 0$

$\Rightarrow f(2^x)(1+x) + f(2^{-x})(1-x) = 0$

Plugging in $x = 1$ gets us: $f(2)(1+1) + f({1 \over 2}) (1-1) = f(2)*2 + 0 = 0$

$\Rightarrow 2*f(2) = 0$ $\Rightarrow f(2) = 0$

(dunno, why I can't post this anymore as a solution)

Alisa Meier - 5 years, 10 months ago

sAme Way!!!

Kaustubh Miglani - 5 years, 2 months ago

Did the same

Aditya Kumar - 5 years ago

Som Ghosh
Jul 18, 2015

$f({ 2 }^{ x })+xf({ 2 }^{ -x })\quad =\quad 1\\ putting\quad -x\quad in\quad place\quad of\quad x\\ f({ 2 }^{ -x })-xf({ 2 }^{ x })\quad =\quad 1\\ solving\quad these\quad two\quad equation\\ f({ 2 }^{ x })\quad =\quad \frac { 1-x }{ 1+{ x }^{ 2 } } \\ hence,\quad putting\quad x=1\\ f(2)\quad =\quad 0$

I like this because it gives you an explicit definition for the function. Let $2^x=y$ and rearrange: $x=\log_2 y$ , you can express the function as $f(y)=\frac{1-\log_2 y}{1+\log_2 y}$