Question

Find the critical numbers of f, describe the open intervals on which f is increasing or decreasing, and locate all relative extrema.

f(x)= x+3/x^3

Answer 1

How do I solve this? I got f'(x) = 1/x^3 - 3(x+3) / x^4

Answer 2

is that right?

Answer 3

I found -6 as a critical number .. I just need to make sure it is correct...

Answer 4

intervals and extrema? is what I haven't found

Answer 5

critical number is the derivative's zero right?
http://www.wolframalpha.com/input/?i=0%3D1%2Fx%5E3+-+3%28x%2B3%29+%2F+x%5E4
which would be x = -9/2

Answer 6

I put in critical numbers for the derivative and it gave me -6

Answer 7

you want the critical point of f(x)
you entered in the critical point for f'(x)

Answer 8

http://www.wolframalpha.com/input/?i=%28x%2B3%29%2Fx%5E3+critical+point

Answer 9

it's not the critical point for the derivative?

Answer 10

the critical point of the derivative would be the original function's point of inflection.
aren't you looking for the critical point for the original function?

Answer 11

yeah you're right.

Answer 12

http://www.wolframalpha.com/input/?i=relative+extrema+of+f%28x%29+%3D+%28x%2B3%29%2Fx%5E2

Answer 13

is that correct for relative extrema

Answer 14

didn't the original function have a x^3 in the denominator?
http://www.wolframalpha.com/input/?i=relative+extrema+of+f%28x%29+%3D+%28x%2B3%29%2Fx%5E3

Answer 15

Oh my gosh. I'm sorry :( I posted it wrong no wonder we were looking at diferent things

Answer 16

it's x^2.

Answer 17

so the critical point would be -6

Answer 18

lol ok :) it's fine and yep :)

Answer 19

and how do I get the intervals?

Answer 20

to find where the original is increasing or decreasing,
find where the derivative is positive or negative

Answer 21

ok my derivative is f'(x) = 1/x^2 - (2(x+3) )/ x^3

Answer 22

how do I find positive and negative?

Answer 23

which can be simplified to (-x-6)/x^3
then we know that there is critical point at x=-6
and asympote at x=0
so plug in a number less then -6 into the derivative and figure out if it's positive or negative.
then plug in a number between -6 and 0 into the derivative and figure out if it's positive or negative
etc.

Answer 24

-5 gives me 0.008

Answer 25

or you can use graph of the derivative
http://www.wolframalpha.com/input/?i=0%3D1%2Fx%5E2+-+%282%28x%2B3%29+%29%2F+x%5E3
and see that before x=-6 y values are negative
between -6 < x < 0 y values are positive

Answer 26

and yup :) so between -6 < x< 0
it is positive

Answer 27

-7 gives me -002

Answer 28

i meant .002

Answer 29

and x<-6<0 is negative

Answer 30

negative is maximum and positive is minimum correct?

Answer 31

negative means original function is decreasing
positive means original function is increasing

Answer 32

right :)

Answer 33

thank you

Answer 34

you're welcome :)