Let f : R to R be a function defined by f(x) | vedclass

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(C) Given,$f(x) = \frac{4^x}{4^x + 2}$.We observe that $f(1 - x) = \frac{4^{1 - x}}{4^{1 - x} + 2} = \frac{4/4^x}{4/4^x + 2} = \frac{4}{4 + 2 \cdot 4^

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$2$

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(C) Given,$f(x) = \frac{4^x}{4^x + 2}$.We observe that $f(1 - x) = \frac{4^{1 - x}}{4^{1 - x} + 2} = \frac{4/4^x}{4/4^x + 2} = \frac{4}{4 + 2 \cdot 4^x} = \frac{2}{2 + 4^x}$.Also,$f(x) + f(1 - x) = \frac{4^x}{4^x + 2} + \frac{2}{4^x + 2} = \frac{4^x + 2}{4^x + 2} = 1$.Therefore,$f(1 - x) = 1 - f(x)$.We need to evaluate $S = f(\frac{1}{4}) + 2 f(\frac{1}{2}) + f(\frac{3}{4})$.Since $f(\frac{1}{4}) + f(1 - \frac{1}{4}) = f(\frac{1}{4}) + f(\frac{3}{4}) = 1$,we can substitute $f(\frac{3}{4}) = 1 - f(\frac{1}{4})$.Thus,$S = f(\frac{1}{4}) + 2 f(\frac{1}{2}) + (1 - f(\frac{1}{4})) = 1 + 2 f(\frac{1}{2})$.Now,calculate $f(\frac{1}{2}) = \frac{4^{1/2}}{4^{1/2} + 2} = \frac{2}{2 + 2} = \frac{2}{4} = \frac{1}{2}$.Substituting this back,$S = 1 + 2(\frac{1}{2}) = 1 + 1 = 2$.

Let $f : R \to R$ be a function defined by $f(x) = \frac{4^x}{4^x + 2}$. What is the value of $f(\frac{1}{4}) + 2 f(\frac{1}{2}) + f(\frac{3}{4})$?

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