What are the pointsof inflection of $f (x) = x^{3} + 3 x^{2} - (\frac{27}{x^{2}})$ $f(x)=x^3+3x^2 - (27/x^2)$ ?

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What are the pointsof inflection of $f (x) = x^{3} + 3 x^{2} - (\frac{27}{x^{2}})$ $f(x)=x^3+3x^2 - (27/x^2)$ ?

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Answer 1 · 2016-02-11T04:47:21

The points of inflection occur where the second derivative is zero.

First find the first derivative.

$f (x) = x^{3} + 3 x^{2} - (\frac{27}{x^{2}})$ $f (x) = x^3 + 3 x^2 - (27/x^2)$
$f (x) = x^{3} + 3 x^{2} - 27 (x^{- 2})$ $f (x) = x^3 + 3 x^2 - 27 (x^{-2})$
$\frac{d f (x)}{d x} = 3 x^{2} + 3 \cdot 2 x - 27 \cdot (- 2) (x^{- 3})$ ${d f (x)} / {dx} = 3 x^2 + 3 * 2 x - 27*(-2)(x^{-3})$
$\frac{d f (x)}{d x} = 3 x^{2} + 6 x + 54 x^{- 3}$ ${d f (x)} / {dx} = 3 x^2 + 6 x + 54 x^{-3}$
or $\frac{d f (x)}{d x} = 3 x^{2} + 6 x + (\frac{54}{x^{- 3}})$ ${d f (x)} / {dx} = 3 x^2 + 6 x + (54 / {x^{-3}})$

Now the second.
$\frac{d^{2} f (x)}{{d x}^{2}} = 3 \cdot 2 x^{1} + 6 \cdot 1 \cdot x^{0} + 54 \cdot (- 3) (x^{- 4})$ ${d^2 f (x)} / {dx^2} = 3*2 x^1 + 6*1*x^0 +54*(-3) (x^{-4})$
$\frac{d^{2} f (x)}{{d x}^{2}} = 6 x + 6 - 162 x^{- 4}$ ${d^2 f (x)} / {dx^2} = 6x + 6 -162 x^{-4}$
set this equal to zero.
$0 = 6 x + 6 - 162 x^{- 4}$ $0 = 6x + 6 -162 x^{-4}$
Multiply both sides by $x^{4}$ $x^4$ ( allowed as long as $x \neq 0$ $x != 0$ and since the function blows up at zero, this is fine).
$0 = 6 x^{5} + 6 x^{4} - 162$ $0 = 6x^5 + 6 x^4 -162$
Divide through by 6!
$0 = x^{5} + x^{4} - 27$ $0 = x^5 + x^4 - 27$ Go to a equation solver (like Maple, Mathcad or Matlab ) and find the 0's.

Check these (probably five) values in the function and the derivative to make sure they aren't doing anything foolish.

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What are the pointsof inflection of $f (x) = x^{3} + 3 x^{2} - (\frac{27}{x^{2}})$ $f(x)=x^3+3x^2 - (27/x^2)$ ?

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What are the pointsof inflection of $f (x) = x^{3} + 3 x^{2} - (\frac{27}{x^{2}})$ $f(x)=x^3+3x^2 - (27/x^2)$ ?