Question

locate the absolute extrema of the function on the closed interval
f(x) = 2x/(x^2+1), [-2,2] if someone could help with the process, it'd be great since I may have gone somewhere wrong

Answer 1

take the derivative, set it equal to zero and solve
then check the zeros of the derivative, also the endpoints of the interval
largest is max, smallest is min

Answer 2

yeah I understand that but I got a really weird answer when I tried to find the derivative

Answer 3

I used the quotient rule and got x = sqrt (-1/3)

Answer 4

after making the derivative equal to 0

Answer 5

i think your derivative is wrong
problem was cooked so that it is nice and easy to find the zeros

Answer 6

\[f'(x)=\frac{-2(x^2-1)}{(x^2+1)^2}\] zeros are easy and easy to check too i think

Answer 7

the derivative that I got before trying to find the 0 value was (6x^2+2)/(X^2+1)^2

Answer 8

I know I'm asking for a lot but is there any way that you could guide me through the process so that I can see what I did wrong?

Answer 9

numerator should be
\[2(x^2+1)-2x(2x)\]

Answer 10

yep thats what I got in the numerator

Answer 11

multiply out and get
\[2x^2+2-4x^2=-2x^2+2=-2(x^2-1)\]

Answer 12

so zeros are \(-1,1\) both of which are in the interval

Answer 13

i am guessing you made an algebra mistake between
\[2(x^2+1)-2x(2x)\]and
\[-2(x^2-1)\]

Answer 14

yeah you're definitely correct, instead of subtracting the two I added, I got it mixed up with the product rule

Answer 15

I understand this fully thanks for taking the time to explain it to me :)