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Sketching a graph based on certain conditions...

by BlackMamba
Tags: based, conditions, graph, sketching
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BlackMamba
#1
Nov27-05, 05:15 PM
P: 187
Hello,

I'm supposed to sketch a graph of f based on condtions I'm given. However some of the conditions I'm given I'm not sure exactly what is supposed to happen. A little help would be greatly appreciated.

Here are the condtions, some of which I think I know what f is supposed to do, some I do not:

[itex]f'(1) = f'(-1) = 0[/itex] : Does this mean there are horizontal asymptotes at y = 1 and y = -1?
[itex]f'(x) < 0[/itex] if [itex]|x| < 1[/itex] : I believe f is decreasing here on (-1, 1)
[itex]f'(x) > 0[/itex] if [itex]1 < |x| < 2[/itex] : I believe f is increasing here on (-2, -1) and (1, 2)
[itex]f'(x) = -1[/itex] if [itex]|x| > 2[/itex] : I don't have any guesses for this one.
[itex]f''(x) < if -2 < x < 0[/itex] : I believe f is concave down on (-2, 0)
inflection point (0, 1) : I believe concavity changes at this point.

Any help would be greatly appreciated.
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BlackMamba
#2
Nov27-05, 10:32 PM
P: 187
Any suggestions at all would be greatly appreciated. I'm confident that I can draw the final graph, but it's just determining what f is doing based on the conditions given of f'(x) and f''(x).

Thanks in advance.
HallsofIvy
#3
Nov28-05, 06:38 AM
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Quote Quote by BlackMamba
Hello,
I'm supposed to sketch a graph of f based on condtions I'm given. However some of the conditions I'm given I'm not sure exactly what is supposed to happen. A little help would be greatly appreciated.
Here are the condtions, some of which I think I know what f is supposed to do, some I do not:
[itex]f'(1) = f'(-1) = 0[/itex] : Does this mean there are horizontal asymptotes at y = 1 and y = -1?
Not asymptotes but horizontal tangent lines at x= 1 and x= -1. You don't know what f(1) and f(-1) are.

[itex]f'(x) < 0[/itex] if [itex]|x| < 1[/itex] : I believe f is decreasing here on (-1, 1)
Yes, and so you now know that f(-1)>= f(1).

[itex]f'(x) > 0[/itex] if [itex]1 < |x| < 2[/itex] : I believe f is increasing here on (-2, -1) and (1, 2)
Yes, and so there is a local maximum at x=-1, a local minimum at x= 1

[itex]f'(x) = -1[/itex] if [itex]|x| > 2[/itex] : I don't have any guesses for this one.[/tex]
This is the most specific one of all! Since f'(x) is a constant for |x|> 2, y= f(x) is a line with slope -1 for x< -2 and x> 2.

[itex]f''(x) < if -2 < x < 0[/itex] : I believe f is concave down on (-2, 0)
inflection point (0, 1) : I believe concavity changes at this point.
Any help would be greatly appreciated.
Yes, the curve is concave down between -2 and 0 and concave up between 0 and 2. Your graph should be a straight line with slope -1 for x< -2, then an "s" curve going up to a maximum at x= -1 then down to a minimum at x= 1, then up to x= 2 where the graph changes to a straight line with slope -1.
The only y-value you are given is that f(0)= 1 since we are told that there is an inflection point at (0,1). There are, of course, an infinite number of graphs, y= f(x), that satisfy these conditions.

BlackMamba
#4
Nov28-05, 08:38 AM
P: 187
Sketching a graph based on certain conditions...

Thanks again, HallsofIvy. I appreciate you giving the extra explanations. They help me understand these concepts a little more.


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