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

Question

William Abdalla · Accepted Answer

The function is convex on #(-oo,-3)uu(0,oo)#.
The function is concave on #(-3,0)#.

First, find the second derivative. First Derivative Use product rule. #f'(x)=(2x-1)e^x+(x^2-x)e^x# #=>e^x(x^2+x-1)# Second Derivative Use product rule again. #f''(x)=e^x(x^2+x-1)+e^x(2x+1)# #=>e^x(x^2+3x)=xe^x(x+3)# Create a sign chart to find when #f''(x)# is positive (convex) and negative (concave). To find the important values on the chart, set #f''(x)=0#. #xe^x(x+3)=0# #x=-3,0# #color(white)(ssssssssss)-3color(white)(ssssssssssssss)0# #larr-------------rarr# #color(white)(sssss)+color(white)(ssssssssssss)-color(white)(ssssssssssss)+# The function is convex on #(-oo,-3)uu(0,oo)#. The function is concave on #(-3,0)#. graph{e^x(x^2-x) [-10, 10, -5, 5]}

Evelyn Agard · Answer

undefined To determine the concavity or convexity of $ f(x) = (x^2 - x)e^x $, we need to find its second derivative and analyze its sign. First derivative: $ f'(x) = (2x - 1)e^x + (x^2 - x)e^x $ Second derivative: $ f''(x) = (2 - 2x)e^x + (2x - 1)e^x + (x^2 - x)e^x $ Simplify the second derivative: $ f''(x) = (x^2 + x)e^x $ Now, to determine concavity or convexity, we look at the sign of $ f''(x) $: - $ f''(x) > 0 $ for $ x < -1 $ or $ x > 0 $. Therefore, the function is convex on $ (-\infty, -1) $ and $ (0, \infty) $. - $ f''(x) < 0 $ for $ -1 < x < 0 $. Therefore, the function is concave on $ (-1, 0) $.