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

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(C) Given $f(x) = \frac{1}{\sqrt{x + 2 \sqrt{2x - 4}}} + \frac{1}{\sqrt{x - 2 \sqrt{2x - 4}}}$.Substitute $x = 11$:$f(11) = \frac{1}{\sqrt{11 + 2 \sqr

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$\frac{6}{7}$

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(C) Given $f(x) = \frac{1}{\sqrt{x + 2 \sqrt{2x - 4}}} + \frac{1}{\sqrt{x - 2 \sqrt{2x - 4}}}$.Substitute $x = 11$:$f(11) = \frac{1}{\sqrt{11 + 2 \sqrt{2(11) - 4}}} + \frac{1}{\sqrt{11 - 2 \sqrt{2(11) - 4}}}$$f(11) = \frac{1}{\sqrt{11 + 2 \sqrt{18}}} + \frac{1}{\sqrt{11 - 2 \sqrt{18}}}$Since $\sqrt{18} = 3 \sqrt{2}$,we have:$f(11) = \frac{1}{\sqrt{11 + 6 \sqrt{2}}} + \frac{1}{\sqrt{11 - 6 \sqrt{2}}}$Note that $11 + 6 \sqrt{2} = 9 + 2 + 2(3)(\sqrt{2}) = (3 + \sqrt{2})^2$ and $11 - 6 \sqrt{2} = (3 - \sqrt{2})^2$.Thus,$f(11) = \frac{1}{3 + \sqrt{2}} + \frac{1}{3 - \sqrt{2}}$$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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