Question

how do you do this? http://puu.sh/7wUfA.png

Answer 1

Looks like a test or hw problem. http://openstudy.com/code-of-conduct

Answer 2

ok then I retract my criticisms

Answer 3

That question isn't too difficult if you are interested in Calculus anyway...

Answer 4

my question is

Answer 5

is the domain

Answer 6

No, I remember getting this question of the AP exam a few years ago lol.

Answer 7

it's not

Answer 8

[-4,2) U (2,0)

Answer 9

I remember it on my first calc 1 test

Answer 10

is that correct?

Answer 11

ok that's good

Answer 12

Why don't you think it exists at 2?

Answer 13

because it's not decr/incr where f'(x) is 0

Answer 14

....it's not zero at 2?

Answer 15

-2*

Answer 16

I think the domain is from -4 to +4. Since the derivative is defined, doesn't that make the function defined (because you could in theory integrate it)?

Answer 17

the question a is where f(x) is decr

Answer 18

f(x) is decreasing whenever f'(x) is less than zero.

Answer 19

Which is [-4,-2)U(-2,0)

Answer 20

thanks!

Answer 21

Do you know why?

Answer 22

I do

Answer 23

thank you for helping

Answer 24

Good. Just checking. :)
And you're welcome.

Answer 25

actually function is decreasing on [-4,0). f'(-2) = 0 only means the tangent line is 0.

Consider the function -x^3. It's a decreasing function on (-inf,inf) even though at x = 0, the tangent line is horizontal.

Answer 26

yes, but when the derivative is 0 it isn't decreasing or increasing.

Answer 27

I agree, it's still neither increasing or decreasing when the derivative equals 0. Not always decreasing.

Answer 28

Hint: Sour is correct

Answer 29

doesn't that mean that the function is not decreasing?

Answer 30

on -2?

Answer 31

a point can not be increasing nor decreasing. Say f'(-1) = -2, this only means the tangent at -1 is -2 but it doesn't makes sense to say the function is decreasing at the point. We need other second x value to compare

Answer 32

But it can be both ;D

Answer 33

at x = 0 is where the function has local minimum because f'(x) changes from negative to positive

Answer 34

This is the graph. It is not decreasing when f'(x)=0. I don't know how you guys learned your way, but it is wrong.

Answer 35

that is wrong^