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anonymous
 5 years ago
Find the critical points for f(t) = t/(1 + t^2)
anonymous
 5 years ago
Find the critical points for f(t) = t/(1 + t^2)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you do the derivative of the equation? Set that equal to zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Also identify points that will make the derivative undefined.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Get your complete answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am stuck on finding the crtitical points from (12t^2(1+t^2)^1). I know that the factored out piece (1 + t^2)^1 is where the derivative is undefined...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let me check something. That isn't the derivative that I ended up with.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should end up with (1t^2)/[(1+t^2)^2\

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then your critical points should be pretty easy to find from there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Was your f'(t) = (1+t^2)^1  2t^2(1+t^2)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am not sure how you came to your derivative.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(t) = t/(1+t ^{2})\] define \[g(t) = t\] \[h(t) = 1 + t ^{2}\] Quotient Rule: \[f'(x) = [h(x)g'(x)  g(x)h'(x)]/[h(x)]^{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can also use the product rule, of course. This is just an easy way to annotate online.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you use the product rule, remember the chain rule with the negative exponent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got \[f'(x) = ((1+t^2)2t^2)/(1+t^2)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Correct. Simplify the numerator.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So now I have \[2t^2/(1+t^2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh! I know what I did the actual answer is \[(1t^2)/(1+t^2)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, what values of t make this 0 or undefined?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No value can make this undefined because t is squared in the denominator...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No real value, right. i will, if you are allowed complex solutions. I'm guessing not, however.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can't multiply the denominator on each side to get rid of it, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can, but it isn't needed. You have done all the steps you need.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But the critical points are when I set f' equal to zero...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A number line would be unrealistic...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The derivative is 0 when \[1  t ^{2} = 0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ohwe only set the numerator equal to zero? Does that apply for all fraction derivatives...because I know that if I set the denominator equal to zero, it would show the values where the function is undefined.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right. If the numerator is 0, the fraction is 0. If the denominator is 0, the fraction is undefined. There are some interesting things that happen if both are 0, but you'll learn those later :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So I have my critical points as +/ 1. This means that one of them can be a global max/min when I test the critical points.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0They can be global, local, or neither.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks so much for helping me and staying online this entire time! I really appreciate it!
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