What are the pointsof inflection of f(x)=x^2 - (27/x^2)?

Feb 19, 2018

$\setminus q \quad \text{the points of inflection are:} \setminus q \quad \left(- 3 , 6\right) , \setminus \quad \left(3 , 6\right) .$

Explanation:

$\text{First, we need to find" \ \ f' '(x) ".}$

$\text{To aid in this, let's rewrite" \ \ f(x) \ \ "to prepare it for}$
$\text{differentiation, and then differentiate:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f \left(x\right) \setminus = \setminus {x}^{2} - \frac{27}{x} ^ 2.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad f \left(x\right) \setminus = \setminus {x}^{2} - 27 {x}^{- 2}$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f ' \left(x\right) \setminus = \setminus 2 x - 27 \left(- 2 {x}^{- 3}\right) .$

$\text{No need to stop and simplify here, we just continue}$
$\text{differentiating to get" \ \ f' '(x) ":}$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus \quad f ' ' \left(x\right) \setminus = \setminus 2 - 27 \left(\left(- 2\right) \setminus \cdot \left(- 3\right) {x}^{- 4}\right)$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus \quad f ' ' \left(x\right) \setminus = \setminus 2 - \left(27\right) \left(6\right) {x}^{- 4} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \left(1\right)$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus \quad f ' ' \left(x\right) \setminus = \setminus 2 - \frac{27 \setminus \cdot 6}{x} ^ 4 .$

$\text{There may be no need to multiply out the numbers -- as we are}$
$\text{going to use" \ \ f''(x) \ \ "in an equation set to 0. In fact, it may be}$
$\text{better not to multiply them out, as we may just have to factor}$
$\text{the result all over again, later, when simplifying the numerical}$
$\text{quantitites -- this idea will be illustrated as we continue below.}$

$\text{Now for the points of inflection, we solve:" \ \ f''(x) = 0 \ \ ":}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad 2 - \frac{27 \setminus \cdot 6}{x} ^ 4 \setminus = \setminus 0$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad 2 \setminus = \setminus \frac{27 \setminus \cdot 6}{x} ^ 4$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad {x}^{4} \setminus = \setminus \frac{27 \setminus \cdot \textcolor{red}{\cancel{6}} 3}{\textcolor{red}{\cancel{2}}}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad {x}^{4} \setminus = \setminus 27 \setminus \cdot 3 \setminus = \setminus {3}^{3} \setminus \cdot 3 \setminus = \setminus {3}^{4}$

$\setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus {x}^{4} \setminus = \setminus {3}^{4}$

$\setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus {x}^{2} \setminus = \setminus \setminus \pm {3}^{2} \setminus = \setminus - {3}^{2} , \setminus \quad {3}^{2}$

$\setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad x \setminus = \setminus \setminus \pm 3 i , \setminus \quad \setminus \pm 3.$

$\text{Keep only the real solutions:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad x \setminus = \setminus - 3 , \setminus \quad 3. \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \left(2\right)$

$\text{These are the only possible points of inflection. To check if}$
$\text{they are actual points of inflection, we can use the Third}$
$\text{Derivative Test. From eqn. (1) above, we had:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f ' ' \left(x\right) \setminus = \setminus 2 - \left(27\right) \left(6\right) {x}^{- 4} .$

$\text{So:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f ' ' ' \left(x\right) \setminus = \setminus 0 - \left(27\right) \left(6\right) \left(- 4\right) {x}^{- 5}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f ' ' ' \left(x\right) \setminus = \setminus - \frac{\left(27\right) \left(6\right) \left(- 4\right)}{x} ^ \left\{5\right\}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f ' ' ' \left(x\right) \setminus = \setminus \frac{\left(27\right) \left(6\right) \left(4\right)}{x} ^ \left\{5 .\right\}$

$\text{Again, no need to multiply out to apply the Third Derivative}$
$\text{Test; all we want to see is if the Third Derivative comes}$
$\text{out non-zero for the possible points of inflection we found in}$
$\text{eqn. (2)} :$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f ' ' ' \left(- 3\right) \setminus = \setminus \frac{\left(27\right) \left(6\right) \left(4\right)}{- 3} ^ \left\{5\right\} \ne 0$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f ' ' ' \left(3\right) \setminus q \quad \setminus \setminus = \setminus \frac{\left(27\right) \left(6\right) \left(4\right)}{3} ^ \left\{5\right\} \ne 0$

$\text{As the Third Derivatives come out non-zero for both solutions}$
$\text{in (3), we conclude that these possible points of inflection are}$
$\text{actual points of inflection.}$

$\text{Thus:}$

$\setminus q \quad \setminus q \quad \setminus \quad \text{the points of inflection occur at:} \setminus q \quad x \setminus = \setminus - 3 , \setminus \quad 3.$

$\text{So, to complete the problem, we need to just calculate the}$
$\text{complete points for these solutions:}$

$f \left(- 3\right) = {\left(- 3\right)}^{2} - \frac{27}{- 3} ^ 2 = 9 - 3 = 6 \setminus \quad \implies$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \text{complete point} = \left(- 3 , 6\right) .$

$f \left(3\right) = {3}^{2} - \frac{27}{3} ^ 2 = 9 - 3 = 6 \setminus \quad \implies \setminus \quad \text{complete point} \setminus = \left(3 , 6\right) .$

$\text{Thus, finally:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{the points of inflection are:} \setminus q \quad \left(- 3 , 6\right) , \setminus \quad \left(3 , 6\right) .$