Question

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

Answer 1

Can you do the derivative of the equation?

Set that equal to zero.

Answer 2

Also identify points that will make the derivative undefined.

Answer 3

thanks

Answer 4

Get your complete answer?

Answer 5

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...

Answer 6

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

Answer 7

okay

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

?

Answer 12

I am not sure how you came to your derivative.

Answer 13

\[f(t) = t/(1+t ^{2})\]

define
\[g(t) = t\]
\[h(t) = 1 + t ^{2}\]

Quotient Rule:
\[f'(x) = [h(x)g'(x) - g(x)h'(x)]/[h(x)]^{2}\]

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

Correct. Simplify the numerator.

Answer 18

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

Answer 19

Not quite

Answer 20

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

Answer 21

There ya go

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

I can graph it...

Answer 27

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

Answer 28

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

Answer 29

A number line would be unrealistic...

Answer 30

The derivative is 0 when

\[1 - t ^{2} = 0\]

Answer 31

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.

Answer 32

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

Answer 33

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.

Answer 34

They can be global, local, or neither.

Answer 35

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

Answer 36

My pleasure