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Roberto Lam
Prof.: M. Hayasaka
Calculus I ePortfolio Project (Mat 201.1670)
In this project we were asked to describe and justify some interesting
features of the following function:
f(x) = x e
(-x 2)
First the graph of this function is the following:
(This graph was first obtained by using Maple 9.5)
The following observations were made:
I. The graph is symmetric.
To verify this f(x) must be –f(x) when x= -x
f(-x)= -xe-(-x)^2
f(-x)= -xe-x^2, so f(-x)=-f(x), meaning that the f(x) is an odd function and a
symmetric function.
II. The approximate value of the local maximum/global maximum point
is x=0.71 ; y=0.43
III. The approximate value of the local minimum/global minimum point
is x=-0.71; y=-0.43
(These approximate values were obtain by looking at the graph by Maple)
To show these algebraically we must find the first derivative of f(x) and
set it to be 0
f’(x) = d/dx (xe-x^2)
= d/dx (x)* e-x^2 + d/dx (e-x^2)*x
f’(x) = e-x^2 -2x2 e-x^2
Set f’(x)=0
e-x^2 -2x2 e-x^2=0
e-x^2(1-2x2)=0
x=+ 1 / 2 , so these are the critical values, so replacing them on f(x) we
find the critical points
f(+
1 / 2 )=(+ 1 / 2 )e-(+
1 / 2 )^2
= +0.42882
so critical points are ( 1 / 2 ,f(+ 1 / 2 )) ; (- 1 / 2 ,f(- 1 / 2 ))
IV. By observing the graph using Maple:
 f(x) is decreasing when x<-0.71
 f(x) is increasing when -0.71<x<0.71
 f(x) is decreasing when x>0.71
To verify this, here is the following table
X
X<- 1 / 2
+ 1/ 2
- 1 / 2 <X<+ 1 / 2
x> 1 / 2
f’(x)
Is negative
Is positive
Is negative
f(x)
Is decreasing
Is increasing
Is
decreasing
0
V. Looking at the graph, it seems like the x-axis is an asymptote.
To show this; lim f(x)=0
x+

so, lim f(x)= lim xe-x^2
x+

x+

= x/ ex^2
lim f(x)= 0, then the x-axis is an asymptote.
x+

VI. By seeing the graph using Maple:
 The graph is concave up when 0>x>-1.20 and x> 1.20
 The graph is concave down when 0<x<1.20 and x<-1.20
To verify this, we have to find the inflection values and then the inflection
points.
Inflection points mean the points where the graph changes its concavity.
To find the Inflection Values
f”(x)= d/dx(e-x^2 -2x2 e-x^2)
= -6xe-x^2 + 4x3 e-x^2
f”(x)= 2x e-x^2(2x2-3)
Set f”(x)=0
2x e-x^2(2x2-3)=0
When x=0 ; x= + 3 / 2 the graph changes its concavity.
To show the concavity (concave up or down), here is the following graph
x
0<x< 3 / 2
0>x>- 3 / 2
x> 3 / 2
X<- 3 / 2
f”(x)
Is negative
Is positive
Is positive
Is negative
f(x)
Is concave
down
Is concave
up
Is concave
up
Is concave
down
VII. And finally here is the graph of the derivative of the function
f(x) = x e
(-x 2)
We can verify the observations first made comparing them with this
graph, the derivative of f(x).
2
d
(-x 2)
2 (-x )
=e
-2 x e
dx