On-line Math 21

On-line Math 21

4.5  Curve Sketching

Example 7
y = Öx-   ___
Öx-1
 
.

Solution


   Asymptotes

Here there are no vertical asymptotes, even though there is an ``end'' to the domain. This function is only defined for x ³ 1 , so that the stuff inside the square roots is always positive.

Now, for horizontal asymptotes,

lim
x® ¥ 
Öx-   ___
Öx-1
 
=

lim
x® ¥ 
æ
è
Öx-   ___
Öx-1
 
ö
ø
æ
è
Öx+   ___
Öx-1
 
ö
ø

æ
è
Öx+   ___
Öx-1
 
ö
ø
=

lim
x® ¥ 
( x-(x-1))
æ
è
Öx+   ___
Öx-1
 
ö
ø
=

lim
x® ¥ 
1
æ
è
Öx+   ___
Öx-1
 
ö
ø
=
0,
so the x -axis is a horizontal asymptote. Since f(x) > 0 , the tail is above the horizontal asymptote on the right (there is only the asymptote on the right).


   Intercepts

Here we should also worry about endpoints. Since x ³ 1 is the domain, we need to find f(1) , which is of course 1 . The only place to look for intercepts is on the x -axis, which requires a solution of
0
=
f(x)
=
Öx-   ___
Öx-1
 
, or
Öx
=
  ___
Öx-1
 
,
which has no solutions. So there are no intercepts.


   Increasing/decreasing, and critical points


f¢(x)
=
1
2Öx
- 1
2   ___
Öx-1
 
=
  ___
Öx-1
 
-Öx

2Öx   ___
Öx-1
 
,
so, just as there were no x -intercepts, there can't be any critical points. f¢(x) < 0 for all x.


   Concavity, and inflection points


f¢¢(x)
=
æ
ç
ç
ç
è
1
2Öx
- 1
2   ___
Öx-1
 
ö
÷
÷
÷
ø
¢
=
æ
ç
è
1
2
x-1/2- 1
2
(x-1)-1/2 ö
÷
ø
¢
=
- 1
4
x-3/2+ 1
4
(x-1)-3/2
=
x3/2-(x-1)3/2
4x3/2(x-1)3/2
,
which is always positive. So, there are no inflection points, and the graph is always concave up.


   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, 00:44.