If f(x) = frac{x}{sqrt{1 + x^2}},then (fofof) | vedclass

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(B) Given $f(x) = \frac{x}{\sqrt{1 + x^2}}$.First,calculate $(fof)(x) = f(f(x)) = f\left(\frac{x}{\sqrt{1 + x^2}}\right)$.$= \frac{\frac{x}{\sqrt{1 +

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$\frac{x}{\sqrt{1 + 3x^2}}$

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(B) Given $f(x) = \frac{x}{\sqrt{1 + x^2}}$.First,calculate $(fof)(x) = f(f(x)) = f\left(\frac{x}{\sqrt{1 + x^2}}\right)$.$= \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{1 + \left(\frac{x}{\sqrt{1 + x^2}}\right)^2}} = \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{1 + \frac{x^2}{1 + x^2}}} = \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{\frac{1 + x^2 + x^2}{1 + x^2}}} = \frac{x}{\sqrt{1 + 2x^2}}$.Now,calculate $(fofof)(x) = f((fof)(x)) = f\left(\frac{x}{\sqrt{1 + 2x^2}}\right)$.$= \frac{\frac{x}{\sqrt{1 + 2x^2}}}{\sqrt{1 + \left(\frac{x}{\sqrt{1 + 2x^2}}\right)^2}} = \frac{\frac{x}{\sqrt{1 + 2x^2}}}{\sqrt{1 + \frac{x^2}{1 + 2x^2}}} = \frac{\frac{x}{\sqrt{1 + 2x^2}}}{\sqrt{\frac{1 + 2x^2 + x^2}{1 + 2x^2}}} = \frac{x}{\sqrt{1 + 3x^2}}$.

If $f(x) = \frac{x}{\sqrt{1 + x^2}}$,then $(fofof)(x) = $

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