"We will need to find where" \ \ f''(x) \ \ "is positive, and where it"
"is negative."
"Recalling the basic theory on concavity, we have:"
\qquad \qquad \qquad f''(x) > 0 \quad rArr \quad "the graph of" \ f(x) \ "is concave up."
\qquad \qquad \qquad f''(x) < 0 \quad rArr \quad "the graph of" \ f(x) \ "is concave down."
"[I apologize, I don't know the language of concave/convex with"
"respect to the concavity of a curve. The language I am"
"familiar with is (concave up)/(concave down). I hope what I"
"can provide to you helps !!]"
"Ok, so let's compute" \ \ f''(x). "We start with:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad f(x) \ = \ ( x^2 - x )/e^x.
"We can rewrite this a little, to prepare it for differentiation --"
"will help a lot ! :"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ f(x) \ = \ ( x^2 - x )e^-x.
"We have avoided the Quotient Rule !! So, continuing:"
\qquad \qquad \qquad \qquad f'(x) \ = \ ( x^2 - x ) [ e^-x ]' + [ x^2 - x ]' e^-x
\qquad \qquad \qquad \qquad \qquad \qquad \quad = \ ( x^2 - x ) e^-x [ -x ]' + (2 x - 1 ) e^-x
\qquad \qquad \qquad \qquad \qquad \qquad \quad = \ ( x^2 - x ) e^-x ( -1 ) + (2 x - 1 ) e^-x
\qquad \qquad \qquad \qquad \qquad \qquad \quad = \ [ ( x^2 - x ) ( -1 ) + (2 x - 1 ) ] e^-x
\qquad \qquad \qquad \qquad \qquad \qquad \quad = \ [ - x^2 + x + 2 x - 1 ] e^-x
\qquad \qquad \qquad \qquad \qquad \qquad \quad = \ ( - x^2 + 3 x - 1 ) e^-x.
"Thus:"
\qquad \qquad \qquad \qquad \qquad \qquad \quad f'(x) \ = \ ( - x^2 + 3 x - 1 ) e^-x.
"So, onward to" \ \ f''(x):
\quad f''(x) \ = \ ( - x^2 + 3 x - 1 ) [ e^-x ]' + [ - x^2 + 3 x - 1 ]' e^-x
\qquad \quad \ \ \ \ = \ ( - x^2 + 3 x - 1 ) e^-x [ -x ]' + ( - 2 x + 3 ) e^-x
\qquad \quad \ \ \ \ = \ ( - x^2 + 3 x - 1 ) e^-x ( -1 ) + ( - 2 x + 3 ) e^-x
\qquad \quad \ \ \ \ = \ [ ( - x^2 + 3 x - 1 ) ( -1 ) + ( - 2 x + 3 ) ] e^-x
\qquad \quad \ \ \ \ = \ [ x^2 - 3 x + 1 - 2 x + 3 ] e^-x
\qquad \quad \ \ \ \ = \ ( x^2 - 5 x +4 ) e^-x .
"Thus:"
\qquad \qquad \qquad \qquad \qquad \qquad \quad f''(x) \ = \ ( x^2 - 5 x +4 ) e^-x.
"Now we need find where" \ \ f''(x) \ \ "is positive, and where it"
"is negative. So we need to solve:"
\qquad \qquad \qquad \qquad f''(x) > 0 \qquad \qquad \qquad \qquad "and" \qquad \qquad \qquad \qquad f''(x) < 0
\qquad ( x^2 - 5 x +4 ) e^-x > 0 \qquad \qquad "and" \qquad \qquad ( x^2 - 5 x +4 ) e^-x < 0.
"The inequalities above can be solved by the method of test"
"points:"
"Solve:" \qquad \qquad \qquad \qquad \qquad \quad \ \ ( x^2 - 5 x +4 ) e^-x = 0.
\qquad \qquad \qquad \qquad \quad \quad \ \ \ [ ( x^2 - 5 x +4 ) e^-x ] cdot e^{+x} = [ 0 ] cdot e^{+x}
\qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ ( x^2 - 5 x +4 ) ( e^-x cdot e^{+x} ) = 0
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \
( x^2 - 5 x + 4 ) cdot 1 = 0
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x^2 - 5 x + 4 = 0
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ ( x - 1 ) ( x - 4 ) = 0
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ x = 1, 4.
"Intervals to Test:" \qquad \qquad \quad \ \ ( - infty, 1 ), \quad ( 1, 4 ), \quad ( 4, infty).
"Results of Test:" \qquad \qquad \qquad \qquad \qquad \quad "+", \qquad \qquad \quad -, \qquad \quad \quad "+".
"Results for Inequalities:"
\qquad \qquad \qquad \qquad f''(x) > 0 \quad "on": \qquad \ ( - infty, 1 ), \quad ( 4, infty);
\qquad \qquad \qquad \qquad f''(x) < 0 \quad "on": \qquad \ ( 1, 4 ).
"Results for Graph of" \ \ f(x):
\qquad \qquad \ "graph of" \ \ f(x) \quad "concave up on": \qquad \ ( - infty, 1 ), \quad ( 4, infty);
\qquad \qquad \ "graph of" \ \ f(x) \quad "concave down on": \qquad \qquad \qquad ( 1, 4 ).
"These are our desired results."
"Summarizing:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad f(x) \ = \ ( x^2 - x )/e^x.
\qquad \qquad \ "graph of" \ \ f(x) \quad "concave up on": \qquad \ ( - infty, 1 ), \quad ( 4, infty);
\qquad \qquad \ "graph of" \ \ f(x) \quad "concave down on": \qquad \qquad \qquad ( 1, 4 ).
