Question

find the extreme on a closed interval
f(x)=[-2,2]

f'(x)=(-2x^2+2)/((x^2+1)^2)

Answer 1

I mean f(x)=(2x)/(x^2+1)

Answer 2

what is f?

Answer 3

oh lol

Answer 4

\[f'(x)=\frac{2(x^2+1)-2x(2x)}{(x^2+1)^2}=\frac{-2x^2+2}{(x^2+1)^2}\]
good work on f' :)

Answer 5

so x^2+1 is never 0

Answer 6

thanks

Answer 7

is the top ever 0?

Answer 8

\[-2x^2+2=0 => 2=2x^2 => 1=x^2\]

Answer 9

\[x=1,-1\]

Answer 10

no

Answer 11

why do you think that?

Answer 12

well because you can sub nothing for x that would make it zero since it has +2

Answer 13

so x=1 or x=-1 does not work?

Answer 14

no you asked if the top could equal zero

Answer 15

yes
it can be zero when x=1,-1

Answer 16

just solve f'(X)=0 and plug in the solutions

Answer 17

so whats 0=(-2x^2+2)/((x^2+1)^2)

Answer 18

\[-2x^2+2=0\]
so when you solve this you get 2 real solutions

Answer 19

\[x^2-1=0\]
does look better to you?

Answer 20

\[-2(x^2-1)=0 => x^2-1=0\]

Answer 21

the deno is always positive, drop it. and we can get -2x^2+2=0

Answer 22

\[x=\pm 1\]

Answer 23

So -2(x^2-1)=0
so -2=0 which is none but wouldnt -1 squared be imaginary?

Answer 24

\[(-1)^2=1 \]

Answer 25

you are thinking of the square root of -1

Answer 26

oh my bad, it's positive 1 not negative

Answer 27

\[x^2-1=0 => x^2 =1 => x= \pm 1=x=1,-1\]

Answer 28

why will it be negative 1 and +1?

Answer 29

\[x^2-1=0\]
\[(x-1)(x+1)=0\]
\[x-1=0 \text{ or } x+1=0\]
\[x=1 \text{ or } x=-1\]

Answer 30

that's right, sorry for the misunderstanding.

Answer 31

if you don;t like that way either lets go back to the first
\[x^2-1=0 =>x^2=1 => \sqrt{x^2}= \sqrt{1} => |x|= 1 => x=\pm 1\]

Answer 32

so anyways now plug in x= -2,-1,1,2 into f to get the value f(x)

Answer 33

yeah I got that part, can you assist me on another problem pls?