If $x = 3$ find $\frac{x}{3} + \frac{3}{x}$

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Solution

$x = 3 \frac{x}{3} + \frac{3}{x} \frac{3}{3} + \frac{3}{3} 1 + 1 = 2$

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Find the cube of $x + \frac{1}{x}$

x = 1 + 2^{1 / 3} + 2^{2 / 3}

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d y / d x

y = {tan}^{- 1} {(\frac{3 x - x^{3}}{1 - 3 x^{2}})}^{'} \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{3}}

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f (x) = x^{3} + 5 x^{2} - 3 x + 5

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