Need help finding f'(x) from this graph

J

Joey29

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Dec 31, 2005
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hey, I'm having trouble finding f'(x) from this picture. Can someone draw a sketch of it? I appreciate any help. Make fun of my graph all you want, I tried to draw it in paint the best I could : )

Its suppose to be symmetric.

f(x)

juai3c.gif


This is what I think is the derivitive of this function:

juba6b.gif


note: Not drawn to scale

I appreciate any help!!!

Joe
 
G'day, Joe.

A good place to start would be to identify where the slope is zero (relative max and mins) as those x-ordinates will have f'(x) = 0.

The portion of the curve you have shown does not tend towards the supposed asymptote, so it's not particularly relevant for the domain we are given.

Is the curve straight (decreasing/increasing at a constant rate) between the relative extrema or would be be an inflexion point in there?
 
The curve is straight (decreasing and increasing at a constant rate).
 
Then f'(x) is a horizontal line (below the x-axis when f(x) is decreasing or above when f(x) is increasing) in those regions.
 
Oh, so it wouldn't have 2 bumps on the f'(x) because it had 3 bumps on f(x)?
 
Mistake!

Sorry there would be two inflection points in this problem!!
 
The curve COULD be sin(x)-A (from -2pi to pi) which would of course make f'(x) = cos(x)
 
That makes the bumps more smooth!

You can use f(x)=sin(x) and f'(x)=cos(x) as a better guide now. Plotting those x-intercepts should put you on good footing.

Curve going down: f'(x) negative.
Curve going up: f'(x) positive.
 
Thanks Guys

No, I think I've got a handle to this question with the help of you. I appreciate your help!


Joe
 
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