Can anybody guide me on this problem?
thank you.
you mean to say that i am going to substitute the numerator first quantity by y and so it is y^3 then the second quantity would be solely y?
then the denominator as well? pls guide me on this sir so confused to find what does maximum value
Substitute $\displaystyle y = x + \frac{1}{x}$ just like emakarov suggested. So $\displaystyle (x + \frac{1}{x})^3 = y^3$.
Also, $\displaystyle (x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x})$. Substitute $\displaystyle x + \frac{1}{x}$ to obtain
$\displaystyle y^3 = x^3 + \frac{1}{x^3} + 3y \Rightarrow x^3 + \frac{1}{x^3} = y^3 - 3y$
Then you can substitute this into the original f(x).
i made the hint given by richard1234 (x^3 + 1/x^3) ^2
and so i have obtained this
f(x)= (y^3 + y^3 - 3y) / (y^6 - (y^3 - 3y)^2 - 2 - 2)
= (y^3 + y^3 - 3y) / (y^6 - ( y^6 + 6y^4 + 9y^2) - 4
= (y^3 + y^3 - 3y) / 6 y^4 - 9y^2 - 4
= y(2y^2 - 3) / 3y^2 ( 2y^2 - 3) - 4
im stuck here ...
need your help
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