On-line Math 21

On-line Math 21

4.5  Curve Sketching

Example 8
y = xe1/x.

Solution


   Asymptotes

As x® 0 , this function gets a little complicated. For x > 0 , yet x near 0, this one-sided limit

lim
x® 0+ 
xe1/x
is an <a href="hospital.html»indeterminate form</a>, of type 0·¥. In order to convert this to something that l'Hôpital's rule can handle, it needs to be switched to either ¥/¥ or 0/0 . Since I don't want to invert the exponential term, we'll convert to ¥/¥:

lim
x® 0+ 
xe1/x
=

lim
x® 0+ 
e1/x
1/x
=

lim
x® 0+ 
e1/x(-1/x2)
-1/x2
=

lim
x® 0+ 
e1/x
=
¥,
so from the right side, the line x = 0 is a vertical asymptote. However, if you approach from the other side,

lim
x® 0- 
xe1/x
is not an indeterminate form at all, since

lim
x® 0- 
e1/x = 0,
because 1/x is headed to negative infinity, and so

lim
x® 0- 
xe1/x = 0
as well. So, the line x = 0 is a vertical asymptote on one side only, and the graph approaches the origin from the other side (although (0,0) is of course not on the graph).

As x® ±¥, then e1/x® e0 = 1 , and so xe1/x approaches ±¥ as x does. But, for x large, since that exponential term is nearly 1, the graph ought to look like a line, and in fact

lim
x® ¥ 
xe1/x-(x+1) =
lim
x® ¥ 
x( e1/x-(1+1/x)) ,
which is an indeterminate form of type ¥·0 , so converting to 0/0 ,

lim
x® ¥ 
xe1/x-x
=

lim
x® ¥ 
( e1/x-(1+1/x))
1/x
=

lim
x® ¥ 
e1/x(-1/x2)-(-1/x2)
-1/x2
, using l¢Hopital¢s rule
=

lim
x® ¥ 
e1/x-1
=
0,
so that the graph has a slant-asymptote y = x+1 .


   Intercepts

Since 0 is not in the domain, there can be no y -intercept, even though there is a one-sided limit towards a point on the y -axis. For x ¹ 0 , though, there is no solution to f(x) = 0 , so that there are no x -intercepts, either.


   Increasing/decreasing, and critical points


f¢(x)
=
1·e1/x+x·e1/x(-1/x2)
=
e1/x æ
ç
è
1- 1
x
ö
÷
ø
,
so there is a critical point at x = 1 only, with critical value f(1) = e , and the curve is increasing ( f¢(x) > 0 ) for x > 1 , and for x < 0 . The curve is decreasing, f¢(x) < 0 , only when 0 < x < 1 .


   Concavity, and inflection points


f¢¢(x)
=
æ
ç
è
e1/x æ
ç
è
1- 1
x
ö
÷
ø
ö
÷
ø
¢
=
e1/x(-1/x2) æ
ç
è
1- 1
x
ö
÷
ø
+e1/x(+1/x2)
=
e1/x
x2
æ
ç
è
æ
ç
è
1
x
-1 ö
÷
ø
+1 ö
÷
ø
=
e1/x
x3
,
so there are no inflection points, the graph is concave up for x > 0 and concave down for x < 0 .


   Draw the graph

Use this information to draw a fair representation of the graph.

[Each of these items should trigger the appearance of a new drawing with that information added.]

Copyright (c) 2000 by David L. Johnson.


File translated from TEX by TTH, version 2.61.
On 21 Dec 2000, 00:46.