"First, we need to find" \ \ f' '(x) "."
"To aid in this, let's rewrite" \ \ f(x) \ \ "to prepare it for"
"differentiation, and then differentiate:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ x^2 - 27 / x^2.
\qquad \qquad :. \qquad \qquad\qquad \qquad \quad f(x) \ = \ x^2 - 27 x^{ -2 }
\qquad \qquad :. \qquad \qquad\qquad \qquad f'(x) \ = \ 2 x - 27 ( -2 x^{ -3 } ).
"No need to stop and simplify here, we just continue"
"differentiating to get" \ \ f' '(x) ":"
\qquad \qquad :. \qquad \quad f''(x) \ = \ 2 - 27 ( ( -2 ) \cdot ( -3 ) x^{ -4 } )
\qquad \qquad :. \qquad \quad f''(x) \ = \ 2 - ( 27 ) ( 6 ) x^{ -4 } \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad (1)
\qquad \qquad :. \qquad \quad f''(x) \ = \ 2 - { 27 \cdot 6 } / x^4 .
"There may be no need to multiply out the numbers -- as we are"
"going to use" \ \ f''(x) \ \ "in an equation set to 0. In fact, it may be"
"better not to multiply them out, as we may just have to factor"
"the result all over again, later, when simplifying the numerical"
"quantitites -- this idea will be illustrated as we continue below."
"Now for the points of inflection, we solve:" \ \ f''(x) = 0 \ \ ":"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 2 - { 27 \cdot 6 } / x^4 \ = \ 0
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 2 \ = \ { 27 \cdot 6 } / x^4
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x^4 \ = \ { 27 \cdot color{red}cancel{ 6 } 3 } / color{red}cancel{ 2 }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x^4 \ = \ 27 \cdot 3 \ = \ 3^3 \cdot 3 \ = \ 3^4
\qquad \qquad \qquad :. \qquad \qquad \qquad \qquad \ x^4 \ = \ 3^4
\qquad \qquad \qquad :. \qquad \qquad \qquad \qquad \ x^2 \ = \ \pm 3^2 \ = \ - 3^2, \quad 3^2
\qquad \qquad \qquad :. \qquad \qquad \qquad \qquad \quad x \ = \ \pm 3 i, \quad \pm 3.
"Keep only the real solutions:"
\qquad \qquad \qquad :. \qquad \qquad \qquad \qquad \qquad x \ = \ - 3, \quad 3. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (2)
"These are the only possible points of inflection. To check if"
"they are actual points of inflection, we can use the Third"
"Derivative Test. From eqn. (1) above, we had:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad f''(x) \ = \ 2 -( 27 ) ( 6 ) x^{ -4 }.
"So:"
\qquad \qquad \qquad \qquad \qquad \qquad f'''(x) \ = \ 0 -( 27 ) ( 6 ) ( -4 ) x^{ -5 }
\qquad \qquad \qquad \qquad \qquad \qquad f'''(x) \ = \ - { ( 27 ) ( 6 ) ( -4 ) } / x^{ 5 }
\qquad \qquad \qquad \qquad \qquad \qquad f'''(x) \ = \ { ( 27 ) ( 6 ) ( 4 ) } / x^{ 5 .}
"Again, no need to multiply out to apply the Third Derivative"
"Test; all we want to see is if the Third Derivative comes"
"out non-zero for the possible points of inflection we found in"
"eqn. (2)":
\qquad \qquad \qquad \qquad \qquad \qquad f'''( -3 ) \ = \ { ( 27 ) ( 6 ) ( 4 ) } / (-3)^{ 5 } != 0
\qquad \qquad \qquad \qquad \qquad \qquad f'''( 3 ) \qquad \ \ = \ { ( 27 ) ( 6 ) ( 4 ) } / 3^{ 5 } != 0
"As the Third Derivatives come out non-zero for both solutions"
"in (3), we conclude that these possible points of inflection are"
"actual points of inflection."
"Thus:"
\qquad \qquad \quad "the points of inflection occur at:" \qquad x \ = \ - 3, \quad 3.
"So, to complete the problem, we need to just calculate the"
"complete points for these solutions:"
f(-3) = (-3)^2 - 27 / (-3)^2 = 9 - 3 = 6 \quad =>
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad "complete point" = (-3, 6).
f(3) = 3^2 - 27 / 3^2 = 9 - 3 = 6 \quad => \quad "complete point" \ = (3, 6).
"Thus, finally:"
\qquad \qquad \qquad \quad "the points of inflection are:" \qquad (-3, 6), \quad (3, 6).
