Find the asymptote, interval of monotonicity, critical points, the local extreme points, intervals of concavity and inflection point of the following functions. Sketch the graph of each.

Question

A) f(x)=x2ex
B) f(x)=|x2 +x-2|
C) f(x)= x2 -6x
(x+1)2

Answer 1

A) f(x) = x^2 e^x , notice how I indicated exponents?
f ' (x) = x^2 e^x + 2x e^x
= e^x(x^2 + 2x)

f'' (x) = x^2 e^x + 2x e^x + 2x e^x + 2 e^x
= e^x (x^2+ 4x + 2)

here is a graph of your first function
http://www.wolframalpha.com/input/?i=y+%3D+x%5E2+e%5Ex
type in your other equations to get the graphs of them

for your first one:
there is no vertical asymtote, as x ---> large, f(x) --> large
but there is a horizontal asymptote of y = 0
as x ----> -large, f(x) becomes smaller and ---> 0
set f ' (x) = 0
e^x(x^2 + 2x)
e^x = 0 ---> no solution
or x^2 + 2x = 0
x(x+2) = 0
x = 0 or x = -2

close up of above result:
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+e%5Ex%2C++-3+%3C+x+%3C+1

f(0) = 0
f(-2) = 4e^-2 = appr .54

from a graph we have a local max at (-2, .54) and a local min at (0,0)

set f '' (x) = 0
e^x (x^2 + 4x + 2) = 0
e^x= 0 , no solution
or x2+ 4x + 2 = 0
x^2+ 4x + 4 = -2 + 4 , I am using completing the square
(x+2)^2 = 2
x+2 = ± √2
x = -2 ± √2

So two points of inflection, as seen in
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+e%5Ex%2C++-8+%3C+x+%3C+1

All you have to do is enter your other equations in this fantastic webpage.