\
"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 ).