# \ #
# "To answer the question, we want to find where the second" #
# "derivative of the function is positive, and where it is negative." #
# "We are given:" #
# \qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ ( 2 x - 2 )( x - 4 )( 3 x - 3 ). #
# "Taking a second look at the function, we see we can rewrite it as:" #
# \qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ 2 ( x - 1 )( x - 4 ) \cdot 3( x - 1 ) #
# \qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ 6 ( x - 4 )( x - 1 )^2. #
# "This makes it easier to differentiate, starting with the" #
# "Product Rule:" #
# \qquad \quad f'(x) \ = \ 6 \cdot ( ( x - 4 )[ ( x - 1 )^2 ]' + [ ( x - 4 ) ]'( x - 1 )^2 ) #
# \qquad \quad f'(x) \ = \ 6 \cdot ( ( x - 4 )[ 2 ( x - 1 ) \cdot 1 ] + [ 1 ]( x - 1 )^2) #
# \qquad \quad f'(x) \ = \ 6 \cdot (2 ( x - 4 )( x - 1 ) + ( x - 1 )^2 ) #
# "Factor out" \ ( x - 1 ) \ "from the inside:" #
# \qquad \quad f'(x) \ = \ 6 \cdot ( x - 1 ) [ 2 ( x - 4 )+ ( x - 1 )^1 ] #
# \qquad \quad f'(x) \ = \ 6 \cdot ( x - 1 ) [ 2 x - 8 + x - 1 ] #
# \qquad \quad f'(x) \ = \ 6 \cdot ( x - 1 ) [ 3 x - 9 ] \ = \ 6 \cdot ( x - 1 ) \cdot 3 \cdot (x - 3 ) #
# \qquad \quad f'(x) \ = \ 18 ( x - 1 ) (x - 3 ). #
# "Now calculate" \ f''(x)":" #
# \qquad \quad f''(x) \ = \ [ 18 ( x - 1 ) (x - 3 ) ]' #
# \qquad \quad f''(x) \ = \ 18 \cdot [ ( x - 1 ) (x - 3 ) ]' #
# \qquad \quad f''(x) \ = \ 18 \cdot [ ( x - 1 ) (x - 3 )' + ( x - 1 )' (x - 3 ) ] #
# \qquad \quad f''(x) \ = \ 18 \cdot [ ( x - 1 ) \cdot 1 + 1 \cdot (x - 3 ) ] #
# \qquad \qquad \qquad \qquad \quad = \ 18 \cdot [ x - 1 + x - 3 ] \ = \ 18 \cdot ( 2 x - 4 ) #
# \qquad \qquad \qquad \qquad \quad = \ 18 \cdot 2 \cdot ( x - 2) \ = \ 36 \cdot ( x - 2 ) #
# \ #
# :. \qquad \qquad \qquad \qquad \qquad \ f''(x) \ = \ 36 \cdot ( x - 2 ). #
# \ #
# "Now for concavity, we find where" \ f''(x) \ "is positive, and where" #
# "it's negative." #
# \qquad \qquad \qquad \qquad \quad "concave up:" \qquad \quad \ \quad \ 36 \cdot ( x - 2 ) > 0; #
# \qquad \qquad \qquad \qquad \quad "concave down:" \qquad \ 36 \cdot ( x - 2 ) < 0. #
# "There are several ways to solve such inequalities. With the" #
# "expression as simple as it is, probably the easiest method is to" #
# "do it directly."#
# "As 36 is positive, we can divide through both sides of the" #
# "inequalities by 36, leaving us with:" #
# \qquad \qquad \qquad \qquad \quad "concave up:" \qquad \quad \quad \ \ x - 2 > 0; #
# \qquad \qquad \qquad \qquad \quad "concave down:" \qquad \ x - 2 < 0. #
# "So:" #
# \qquad \qquad \qquad \quad "concave up:" \qquad \quad \quad \ \ x > 2, \quad "or the interval" \quad ( 2, \infty ); #
# \qquad \qquad \qquad \quad "concave down:" \qquad \ x < 2, \quad "or the interval" \quad ( -\infty, 2 ). #
# "This our answer." #
# \ #
# "Summarizing:" #
# \qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ ( 2 x - 2 )( x - 4 )( 3 x - 3 ): #
# \qquad \qquad \quad "is concave up:" \qquad \quad \quad \quad \ \ "on the interval" \quad ( 2, \infty ); #
# \qquad \qquad \quad "is concave down:" \qquad \quad \ "on the interval" \quad ( -\infty, 2 ). #
