Answer:
Sketched & Analyzed.
Step-by-step explanation:
We are given: [tex]f(x)=x^3e^x[/tex]
First derivative of f(x):
[tex]f'(x)=3x^2e^x+x^3e^x=x^2e^x(x+3)[/tex]
There are a root at x = -3 and double roots at x = 0.
So it is decreasing on (-∞, -3) and increasing on (-3, ∞).
Second derivative of f(x):
[tex]f''(x)=e^xx^3+6e^xx^2+6e^xx=xe^x(x^2+6x+6)[/tex]
Roots are [tex]x = 0,\: x = -3-\sqrt3, \:x=-3+\sqrt3[/tex]
So it is concave down on [tex](-\infty,-3-\sqrt3)[/tex] ∪ [tex](-3-\sqrt3,0)[/tex] and concave up on [tex](-3-\sqrt3,-3+\sqrt3)[/tex] ∪ [tex](0,\infty)[/tex]
The sketch is given in the attachment.
