If f:R−{0} → R is defined by f(x)=x+1x and if fk=[f(x)]k for k≥1 then f4(x)−f(x4)−4f2(x)=

f(x)=x+1xf4(x)(x+1x)4=(x+1x)2=(x2+1x2+2)2=x4+1x4+4+2+4x2+4x2=x4+1x4+6+4x2+4x2f4(x)−f(x4)−4f2(x)=x4+1x4+6+4x2+4x2−(x4+1x4)−4(x2+1x2+2)=x4+1x4+6+4x2+4x2−x4−1x4−4x2−4x2−8=−2

If f:R−{0} → R is defined by f(x)=x+1x and if fk=[f(x)]k for k≥1 then f4(x)−f(x4)−4f2(x)=

f(x)=x+1xf4(x)(x+1x)4=(x+1x)2=(x2+1x2+2)2=x4+1x4+4+2+4x2+4x2=x4+1x4+6+4x2+4x2f4(x)−f(x4)−4f2(x)=x4+1x4+6+4x2+4x2−(x4+1x4)−4(x2+1x2+2)=x4+1x4+6+4x2+4x2−x4−1x4−4x2−4x2−8=−2

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$I f f : R - {0} \to R i s d e f i n e d b y f (x) = x + \frac{1}{x} a n d i f f^{k} = {[f (x)]}^{k} f o r k \geq 1 t h e n f^{4} (x) - f (x^{4}) - 4 f^{2} (x) =$

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-1

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Detailed Solution

$\begin{array}{l} f (x) = x + \frac{1}{x} \\ f^{4} (x) {(x + \frac{1}{x})}^{4} = {(x + \frac{1}{x})}^{2} = {(x^{2} + \frac{1}{x^{2}} + 2)}^{2} \\ = x^{4} + \frac{1}{x^{4}} + 4 + 2 + \frac{4}{x^{2}} + 4 x^{2} \\ = x^{4} + \frac{1}{x^{4}} + 6 + \frac{4}{x^{2}} + 4 x^{2} \\ f^{4} (x) - f (x^{4}) - 4 f^{2} (x) = x^{4} + \frac{1}{x^{4}} + 6 + \frac{4}{x^{2}} + 4 x^{2} - (x^{4} + \frac{1}{x^{4}}) - 4 (x^{2} + \frac{1}{x^{2}} + 2) \\ = x^{4} + \frac{1}{x^{4}} + 6 + \frac{4}{x^{2}} + 4 x^{2} - x^{4} - \frac{1}{x^{4}} - 4 x^{2} - \frac{4}{x^{2}} - 8 \\ = - 2 \end{array}$