f'(t) = (1-t^2)/(1+t^2)^2. I found the critical points to be +/- 1. I need to test them so that I can find a local max and local min, so I find a number before and after each point to see if there is a sign change. For some reason, my checks always so that the critical point candidates do not work. Can any one help? My original function is f(t) = t/(1+t^2).

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f'(t) = (1-t^2)/(1+t^2)^2. I found the critical points to be +/- 1. I need to test them so that I can find a local max and local min, so I find a number before and after each point to see if there is a sign change. For some reason, my checks always so that the critical point candidates do not work. Can any one help? My original function is f(t) = t/(1+t^2).

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When I tried -2 and 2 I got -3/25 so it is decreasing before and after. At 0 the derivative equals 1 so it in increasing in between. This means -1 is a relative minimum and 1 is a relative maximum.
Did you use f'(2) and f'(-2) to come to your conclusion about the relative minimum?
What you could do is take the second derivative and find out the concavity if f''(t)>0 you have a local min , if f''(t)<0 you have a local max

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Would I use f''(2)--plug is in to find the sign of the second derivative at that point?
I did use f'(-2) and f'(2)
if you use the second derivative you would use f"(-1) and f"(1)
use the critical points found from the first derivative to find out the concavity
yes Tbates is right
I thought that if the sign is the same on either side of the critical point, the point is undefined. Is that what happens? My answers for f'(2) and f'(-2) are the same sign.
The first derivative tells us the slope of the original function (increasing or decreasing) the second derivative tells us the concavity (concave up or concave down).
With the sign change you want to test points on either side of the critical point. In this case you have two critical points so you have to test points on either side of both of the points. In this case I would use -2 and 0 for points around x = -1 and I would use 0 and 2 for points around x = 1.
That gives me pos-->neg for each interval
sorry, neg-->pos for the first, pos--> pos for the second
The second one should be positive --> negative. At x = 2 the second derivative is -3/25
if you have a change in concavity.... means you have an inflection point
I am using first derivative ... is that wrong?
nope, the first derivative will do the trick for this problem.
I don't see where you got -3/25--my calculator says that f'(2) = .4
(1-2^2)/[(1+2^2)^2] is (1-4)/[(1+4)^2] = -3/(5^2)= -3/25
Looking at the values from the original function! Sorry about that. Now I see. So t = 1 is a local max, and t = -1 is a local minimum.
Exactly!
Thanks so much everyone!

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