If f: R-{0} rightarrow R is defined by f(x)=x | vedclass

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(A) Given $f(x) = x + \frac{1}{x}$.We need to evaluate $f^4(x) - f(x^4) - 4f^2(x)$.First,calculate $f^2(x) = (x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{

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-$2$

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(A) Given $f(x) = x + \frac{1}{x}$.We need to evaluate $f^4(x) - f(x^4) - 4f^2(x)$.First,calculate $f^2(x) = (x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$.Next,calculate $f^4(x) = (f^2(x))^2 = (x^2 + 2 + \frac{1}{x^2})^2 = (x^2)^2 + 2^2 + (\frac{1}{x^2})^2 + 2(x^2)(2) + 2(x^2)(\frac{1}{x^2}) + 2(2)(\frac{1}{x^2}) = x^4 + 4 + \frac{1}{x^4} + 4x^2 + 2 + \frac{4}{x^2} = x^4 + 4x^2 + 6 + \frac{4}{x^2} + \frac{1}{x^4}$.Now,calculate $f(x^4) = x^4 + \frac{1}{x^4}$.Also,$4f^2(x) = 4(x^2 + 2 + \frac{1}{x^2}) = 4x^2 + 8 + \frac{4}{x^2}$.Substitute these into the expression:$f^4(x) - f(x^4) - 4f^2(x) = (x^4 + 4x^2 + 6 + \frac{4}{x^2} + \frac{1}{x^4}) - (x^4 + \frac{1}{x^4}) - (4x^2 + 8 + \frac{4}{x^2})$.$= x^4 + 4x^2 + 6 + \frac{4}{x^2} + \frac{1}{x^4} - x^4 - \frac{1}{x^4} - 4x^2 - 8 - \frac{4}{x^2}$.$= 6 - 8 = -2$.

If $f: R-\{0\} \rightarrow R$ is defined by $f(x)=x+\frac{1}{x}$ and if $f^k(x)=[f(x)]^k$ for $k \geq 1$,then find the value of $f^4(x)-f(x^4)-4f^2(x)$.

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