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(C) Given $f(x) f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right)$.Let $f(x) = ax^2 + bx + c$. The given equation is equivalent to $f(x) -

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$\frac{13}{6}$

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(C) Given $f(x) f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right)$.Let $f(x) = ax^2 + bx + c$. The given equation is equivalent to $f(x) - 1 = \frac{1}{f(1/x) - 1}$ or similar forms. $A$ standard solution for this functional equation is $f(x) = x^n + 1$ or $f(x) = -x^n + 1$.Since $f(x)$ is a quadratic function,we take $f(x) = x^2 + 1$.Calculating the values:$f\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^2 + 1 = \frac{4}{9} + 1 = \frac{13}{9}$.$f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 + 1 = \frac{9}{4} + 1 = \frac{13}{4}$.Now,$\sqrt{f\left(\frac{2}{3}\right) + f\left(\frac{3}{2}\right)} = \sqrt{\frac{13}{9} + \frac{13}{4}} = \sqrt{\frac{52 + 117}{36}} = \sqrt{\frac{169}{36}} = \frac{13}{6}$.

If $f(x)$ is a quadratic function such that $f(x) f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right)$,then $\sqrt{f\left(\frac{2}{3}\right) + f\left(\frac{3}{2}\right)} = $

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