Question

Find f(2)(0) if f(x)= x^2-1/x^2+2x+1. Is f(0) a relative maximum or a relative minimum of f?

Answer 1

We need to first find the first derivative.
I assume you mean (x^2-1)/(x^2+2x+1)
Though I could be wrong and you actually mean x^2-1/x^2+2x+1

Answer 2

youre correct

Answer 3

First of all we could make this prettier by canceling a factor from numerator in denominator of (x^2-1)/(x^2+2x+1)
we don't have to do this but I think it will make our future with this problem much better.

Answer 4

we cancel x+1?

Answer 5

yep

Answer 6

\[f(x)=\frac{x-1}{x+1}\]
Doesn't this look much easier to find the derivative of?

So what is f'
You need to use what rule?

Answer 7

quotient deriv

Answer 8

yep

Answer 9

so its (x+1)(x) - (x-1)(x) / (x+1)^2

Answer 10

\[f'(x)=\frac{(x-1)'(x+1)-(x+1)'(x-1)}{(x+1)^2} \]
I think you mean derivative of either (x-1) or (x+1) is 1 not x.

Answer 11

ops!

Answer 12

So simplify or f'
then we will continue to go ahead and find the second derivative

Answer 13

im not sure how you simplify this...

Answer 14

You don't know how to simplify (x+1)-(x-1)
?

Answer 15

oh but wat about the denominator?

Answer 16

It is already simplified

Answer 17

i got 2

Answer 18

\[f'(x)=\frac{2}{(x+1)^2} \]

Answer 19

now find the second derivative

Answer 20

(x+1)^2 (0) - (2) 2(x+1) / (x+1)^4

Answer 21

i think

Answer 22

\[f''(x)=\frac{(2)'(x+1)^2-2((x+1)^2)'}{(x+1)^4}\]
You did great.
Now simplify the top.

Answer 23

4x+4

Answer 24

or is it -4x-4

Answer 25

on top you have -2(2)(x+1)

Answer 26

-4(x+1)/(x+1)^4

you can actually reduce this fraction

Answer 27

-4/(x+1)^3

Answer 28

yep that is the second derivative

Answer 29

now surely they don't mean "is f(0) a rel min or max?"
because x=0 is not even a critical number