# "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). #
