anonymous
  • anonymous
Find the critical points for f(t) = t/(1 + t^2)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Can you do the derivative of the equation? Set that equal to zero.
anonymous
  • anonymous
Also identify points that will make the derivative undefined.
anonymous
  • anonymous
thanks

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anonymous
  • anonymous
Get your complete answer?
anonymous
  • anonymous
I am stuck on finding the crtitical points from (1-2t^2(1+t^2)^-1). I know that the factored out piece (1 + t^2)^-1 is where the derivative is undefined...
anonymous
  • anonymous
Let me check something. That isn't the derivative that I ended up with.
anonymous
  • anonymous
okay
anonymous
  • anonymous
You should end up with (1-t^2)/[(1+t^2)^2\
anonymous
  • anonymous
Then your critical points should be pretty easy to find from there.
anonymous
  • anonymous
Was your f'(t) = (1+t^2)^-1 - 2t^2(1+t^2)^2
anonymous
  • anonymous
?
anonymous
  • anonymous
I am not sure how you came to your derivative.
anonymous
  • anonymous
\[f(t) = t/(1+t ^{2})\] 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
  • anonymous
You can also use the product rule, of course. This is just an easy way to annotate online.
anonymous
  • anonymous
If you use the product rule, remember the chain rule with the negative exponent.
anonymous
  • anonymous
I got \[f'(x) = ((1+t^2)-2t^2)/(1+t^2)^2\]
anonymous
  • anonymous
Correct. Simplify the numerator.
anonymous
  • anonymous
So now I have \[-2t^2/(1+t^2)\]
anonymous
  • anonymous
Not quite
anonymous
  • anonymous
Oh! I know what I did-- the actual answer is \[(1-t^2)/(1+t^2)^2\]
anonymous
  • anonymous
There ya go
anonymous
  • anonymous
So, what values of t make this 0 or undefined?
anonymous
  • anonymous
No value can make this undefined because t is squared in the denominator...
anonymous
  • anonymous
No real value, right. i will, if you are allowed complex solutions. I'm guessing not, however.
anonymous
  • anonymous
I can't multiply the denominator on each side to get rid of it, right?
anonymous
  • anonymous
I can graph it...
anonymous
  • anonymous
You can, but it isn't needed. You have done all the steps you need.
anonymous
  • anonymous
But the critical points are when I set f' equal to zero...
anonymous
  • anonymous
A number line would be unrealistic...
anonymous
  • anonymous
The derivative is 0 when \[1 - t ^{2} = 0\]
anonymous
  • anonymous
Oh---we 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
  • anonymous
Right. 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
  • anonymous
So 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
  • anonymous
They can be global, local, or neither.
anonymous
  • anonymous
Thanks so much for helping me and staying online this entire time! I really appreciate it!
anonymous
  • anonymous
My pleasure

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