Find the critical points for f(t) = t/(1 + t^2)

- anonymous

Find the critical points for f(t) = t/(1 + t^2)

- katieb

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- anonymous

Can you do the derivative of the equation?
Set that equal to zero.

- anonymous

Also identify points that will make the derivative undefined.

- anonymous

thanks

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## More answers

- anonymous

Get your complete answer?

- 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

Let me check something. That isn't the derivative that I ended up with.

- anonymous

okay

- anonymous

You should end up with (1-t^2)/[(1+t^2)^2\

- anonymous

Then your critical points should be pretty easy to find from there.

- anonymous

Was your f'(t) = (1+t^2)^-1 - 2t^2(1+t^2)^2

- anonymous

?

- anonymous

I am not sure how you came to your derivative.

- 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

You can also use the product rule, of course. This is just an easy way to annotate online.

- anonymous

If you use the product rule, remember the chain rule with the negative exponent.

- anonymous

I got \[f'(x) = ((1+t^2)-2t^2)/(1+t^2)^2\]

- anonymous

Correct. Simplify the numerator.

- anonymous

So now I have \[-2t^2/(1+t^2)\]

- anonymous

Not quite

- anonymous

Oh! I know what I did-- the actual answer is \[(1-t^2)/(1+t^2)^2\]

- anonymous

There ya go

- anonymous

So, what values of t make this 0 or undefined?

- anonymous

No value can make this undefined because t is squared in the denominator...

- anonymous

No real value, right. i will, if you are allowed complex solutions. I'm guessing not, however.

- anonymous

I can't multiply the denominator on each side to get rid of it, right?

- anonymous

I can graph it...

- anonymous

You can, but it isn't needed. You have done all the steps you need.

- anonymous

But the critical points are when I set f' equal to zero...

- anonymous

A number line would be unrealistic...

- anonymous

The derivative is 0 when
\[1 - t ^{2} = 0\]

- 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

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

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

They can be global, local, or neither.

- anonymous

Thanks so much for helping me and staying online this entire time! I really appreciate it!

- anonymous

My pleasure

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