If f(x) = frac{1}{sqrt{x+2 sqrt{2x-4}}} + fra | vedclass

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(C) We have, $f(x) = \frac{1}{\sqrt{x+2 \sqrt{2x-4}}} + \frac{1}{\sqrt{x-2 \sqrt{2x-4}}}$Substituting $x = 11$:$f(11) = \frac{1}{\sqrt{11+2 \sqrt{2(11

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$\frac{6}{7}$

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(C) We have, $f(x) = \frac{1}{\sqrt{x+2 \sqrt{2x-4}}} + \frac{1}{\sqrt{x-2 \sqrt{2x-4}}}$Substituting $x = 11$:$f(11) = \frac{1}{\sqrt{11+2 \sqrt{2(11)-4}}} + \frac{1}{\sqrt{11-2 \sqrt{2(11)-4}}}$$f(11) = \frac{1}{\sqrt{11+2 \sqrt{18}}} + \frac{1}{\sqrt{11-2 \sqrt{18}}}$$f(11) = \frac{1}{\sqrt{11+6 \sqrt{2}}} + \frac{1}{\sqrt{11-6 \sqrt{2}}}$Since $11+6 \sqrt{2} = 9+2+2(3)(\sqrt{2}) = (3+\sqrt{2})^2$ and $11-6 \sqrt{2} = (3-\sqrt{2})^2$,$f(11) = \frac{1}{3+\sqrt{2}} + \frac{1}{3-\sqrt{2}}$$f(11) = \frac{(3-\sqrt{2}) + (3+\sqrt{2})}{(3+\sqrt{2})(3-\sqrt{2})}$$f(11) = \frac{6}{9-2} = \frac{6}{7}$

If $f(x) = \frac{1}{\sqrt{x+2 \sqrt{2x-4}}} + \frac{1}{\sqrt{x-2 \sqrt{2x-4}}}$ for $x > 2$, then $f(11)$ is equal to

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