For what values of x is #f(x)=(x^2−x)/e^x# concave or convex?

Question

Christopher Allers · Accepted Answer

# \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 ). #

# "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 ). #

Ezra Adams · Answer

undefined To determine where the function $ f(x) = \frac{{x^2 - x}}{{e^x}} $ is concave or convex, we need to find the second derivative of $ f(x) $ and then analyze its sign.

The first derivative of $ f(x) $ is:
$$ f'(x) = \frac{{(2x - 1)e^x - (x^2 - x)e^x}}{{(e^x)^2}} $$

Simplify it to get:
$$ f'(x) = \frac{{(2x - 1 - x^2 + x)e^x}}{{e^{2x}}} = \frac{{(x^2 + x - 1)e^x}}{{e^{2x}}} $$

Now, differentiate $ f'(x) $ to get the second derivative:
$$ f''(x) = \frac{{(2x + 1)(x^2 + x - 1)e^x - (x^2 + x - 1)e^x}}{{e^{2x}}} $$

Simplify it to get:
$$ f''(x) = \frac{{(2x^3 + 3x^2 - 2x - 1)e^x}}{{e^{2x}}} $$

Now, to determine where $ f(x) $ is concave or convex, we need to examine the sign of $ f''(x) $:

If $ f''(x) > 0 $ for a particular $ x $, then $ f(x) $ is convex at that point.
If $ f''(x) < 0 $ for a particular $ x $, then $ f(x) $ is concave at that point.

To find where $ f''(x) = 0 $ (possible inflection points), we set $ f''(x) = 0 $ and solve for $ x $. However, this equation is quite complex and may not have simple solutions. Therefore, numerical methods or graphical analysis may be needed to find the exact values of $ x $ where the function changes concavity.