# Please may someone find the vertex of the graph f(X) = 2(piX^2 +(330/X)).

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### 1 Answer

You need to determine the critical point of the graph of `f(x) = 2(pix^2 +(330/x)), ` hence you need to find the derivative of the function.

Differentiating the function with respect to x yields:

`f'(x) = 4pi*x - 330/(x^2) =gt f'(x) = (4pix^3 - 330)/(x^2)`

You need to find the roots of derivative such that:

`(4pix^3 - 330)/(x^2) = 0 =gt 4pix^3 - 330 = 0`

`` `x^3 - 330/(4pi) = 0`

You need to substitute the difference of cubes by the following product such that:

`(x - root(3)(330/(4pi)))(x^2 + x*root(3)(330/(4pi)) + root(3)(330/(4pi))^2)` =`x^3 - 330/(4pi)`

Hence, `x - root(3)(330/(4pi)) = 0` if `x = root(3)(330/(4pi))`

The equation `x^2 + x*root(3)(330/(4pi)) + root(3) (330/(4pi))^2` = 0 has no real zeroes.

The root of derivative represents the critical value of the function, hence you may find the critical point substituting `root(3)(330/(4pi)) ` for x in the equation of function.

**Hence, the critical value of function is :`f(root(3)(330/(4pi))) = 2pi*(root(3)(330/(4pi))^2) + 330/(root(3)(330/(4pi))).` **