If \(x=\log_4\left(\dfrac{2f(x)}{1-f(x)}\right)\), then find \(f(2010)+f(-2009)\)
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From \(x=\log_4\left(\dfrac{2f(x)}{1-f(x)}\right)\), raising both sides as powers of 4: \(4^x=\dfrac{2f(x)}{1-f(x)}\).
Solving for f(x): \(4^x(1-f(x))=2f(x)\Rightarrow 4^x=f(x)(2+4^x)\Rightarrow f(x)=\dfrac{4^x}{2+4^x}\).
Now compute \(f(1-x)=\dfrac{4^{1-x}}{2+4^{1-x}}\). Multiply numerator and denominator by \(4^x\): \(f(1-x)=\dfrac{4}{2\cdot4^x+4}=\dfrac{2}{4^x+2}\).
Add: \(f(x)+f(1-x)=\dfrac{4^x}{4^x+2}+\dfrac{2}{4^x+2}=\dfrac{4^x+2}{4^x+2}=1\).
So for any x, \(f(x)+f(1-x)=1\). Since \(2010+(-2009)=1\), we have \(-2009=1-2010\), so \(f(-2009)=f(1-2010)\).
Therefore \(f(2010)+f(-2009)=f(2010)+f(1-2010)=1\).
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