Question

Let f(x)=4(x−2)2/3+4. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima).

1. f is increasing on the intervals
2. f is decreasing on the intervals
3. The relative maxima of f occur at x =
4. The relative minima of f occur at x =

Notes: In the first two, your answer should either be a single interval, such as (0,1), a comma separated list of intervals, such as (-inf, 2), (3,4), or the word "none".

Answer 1

Have you considered the 1st Derivative? Please demonstrate.

Answer 2

ok

Answer 3

8/(3 (-2 + x)^(1/3))

Answer 4

thats the derivative

Answer 5

@tkhunny then set it equal to 0 right?

Answer 6

I wasn't getting that at all. Spend a little more time making sure you write what you intend.

\(f(x)=4(x−2)^{2/3}+4\)

\(f'(x)=4(2/3)(x−2)^{-1/3} = (8/3)(x-2)^{-1/3} = \dfrac{8}{3(x-2)^{1/3}}\)

Do we agree?

Answer 7

yes

Answer 8

when it comes to typing, i do my work on paper and i just tell people what i get

Answer 9

You had two different errors in your version. Please be more careful. Also, what possessed you to change (x-2) into (-2+x)?

It does no good to make excuses for poor communication. Add appropriate parentheses. Do what it takes to be clear in all forms of communication.

In this case, I would NOT set this equal to zero and solve. Why do you suppose that would be? Normally, that would be the things to do.

Answer 10

roger that, ill talk and explain more from now on

Answer 11

could you tell me the answers for this? please

Answer 12

Answer my question.

In this case, I would NOT set this equal to zero and solve. Why do you suppose that would be? Normally, that would be the thing to do. Look carefully at the derivative.

Answer 13

the reason why i set the derivative equal to 0, is because someone else explained me that is what i am suppose to do and i have done a few problems like this and it has worked

Answer 14

then i again looking at the derivative i should bring the root to numerator right?

Answer 15

That's why I said that is normally the thing to do. That's very good. Unfortunately, that will not help with this problem. You must see why. Go ahead and try it. See where it leads you.

It makes no difference where the root is written. It means the same thing and produces the same unfortunate result.

Answer 16

@dan815 help

Answer 17

You're not going to answer my question? Did you try to solve for f'(x) = 0?

Answer 18

I did the derivative, i just dont know what to do next

Answer 19

yes

Answer 20

What was your result? Is f'(x) ever zero?

Answer 21

i dont fins a solution

Answer 22

That's because there isn't one. This derivative is never zero. HOWEVER, and this is a really big "however", there may still be a minimum or maximum. IF there is, it will be in some weird place where the derivative does not exist. Can you find a place where the derivative does not exist?

Answer 23

@tkhunny no

Answer 24

I was hoping you would find it. Try x = 2. What is the value of the derivative at x = 2?

Answer 25

infinity

Answer 26

Well, that's sort of correct, but that does not provide a sufficient understanding of what we are seeing. The value of the derivative at x = 2 is NOT "infinity". "Infinity" isn't a number. The technical description we need is that the derivative DOES NOT EXIST at x = 2.

The last thing we need is the realization that the derivative is ALWAYS negative for x < 2 and the derivative is ALWAYS positive for x > 2. This tends to suggest that x = 2 is a global minimum of the original function.

Answer 27

See how the derivative might have a problem at x = 2?

Answer 28

thank you so much, i figured the rest out @tkhunny