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anonymous
 5 years ago
f'(t) = (1t^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).
anonymous
 5 years ago
f'(t) = (1t^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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did you use f'(2) and f'(2) to come to your conclusion about the relative minimum?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Would I use f''(2)plug is in to find the sign of the second derivative at that point?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I did use f'(2) and f'(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you use the second derivative you would use f"(1) and f"(1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use the critical points found from the first derivative to find out the concavity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0With 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That gives me pos>neg for each interval

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, neg>pos for the first, pos> pos for the second

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The second one should be positive > negative. At x = 2 the second derivative is 3/25

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you have a change in concavity.... means you have an inflection point

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am using first derivative ... is that wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope, the first derivative will do the trick for this problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't see where you got 3/25my calculator says that f'(2) = .4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(12^2)/[(1+2^2)^2] is (14)/[(1+4)^2] = 3/(5^2)= 3/25

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Looking 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks so much everyone!
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