- #1

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I got 0 and -(8/27) as the criticical numbers.

Am I right?

I also want to know if this function as a max and a min.

Note:2/3 is an exponent.

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- Thread starter realism877
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- #1

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I got 0 and -(8/27) as the criticical numbers.

Am I right?

I also want to know if this function as a max and a min.

Note:2/3 is an exponent.

- #2

- 22,129

- 3,297

Use the x

Your critical points are correct. So can you tell me when the function is increasing and decreasing?? This will give you information about possible maxima and minima.

- #3

SammyS

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Hello realism877.Function:x+x^{2/3}

I got 0 and -(8/27) as the critical numbers.

Am I right?

I also want to know if this function as a max and a min.

Note:2/3 is an exponent.

(Use the ' X

How did you get the critical numbers?

What did you get for a first derivative?

x+x

- #4

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Use the x^{2}and the x_{2}buttons for displaying exponents.

Your critical points are correct. So can you tell me when the function is increasing and decreasing?? This will give you information about possible maxima and minima.

The function is increasing (-infinity, -8/27)u(0,+infinity)

Decreasing (-8/27)

Absolute max= (-8/27, .1481)

Absolute min= (0,0)

Are my findings correct?

- #5

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The function is increasing (-infinity, -8/27)u(0,+infinity)

Decreasing (-8/27)

You mean that it's decreasing in (-8/27,0), right?

Absolute max= (-8/27, .1481)

Absolute min= (0,0)

These certainly are

The only way to know is by calculating the limits in [itex]\pm \infty[/itex] and see to where the function increases...

- #6

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Yes, (-8/27, 0) decreasingYou mean that it's decreasing in (-8/27,0), right?

These certainly arerelativeminima and maxima, but that doesn't make them absolute. the function increases after 0, so it might get very big. The function increases before -8/27, so it might get really small.

The only way to know is by calculating the limits in [itex]\pm \infty[/itex] and see to where the function increases...

Are they local maxima and minima? The values I posted.

- #7

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Yes, (-8/27, 0) decreasing

Are they local maxima and minima? The values I posted.

The function increases before -8/27 and decreases after, so it's a local maximum. The same with local minimum.

But you'll need to do some more work to determine whether they are global maxima/minima.

- #8

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The function increases before -8/27 and decreases after, so it's a local maximum. The same with local minimum.

But you'll need to do some more work to determine whether they are global maxima/minima.

There are none.

The function is (-infinity, +infinity)

Am I on to something?

- #9

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There are none.

The function is (-infinity, +infinity)

Am I on to something?

Yes, you are entirely correct! You might want to prove that the range is [itex](-\infty,+\infty)[/itex]...

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SammyS

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- #11

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Yes, you are entirely correct! You might want to prove that the range is [itex](-\infty,+\infty)[/itex]...

Thanks

I did the second derivative check, and I tried to get the critical numbers. it resulted to 1=0.

From my calculator it looks like there is concavity down. But how can I determine concavity algebraically?

- #12

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How do I show?

- #13

SammyS

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lim_{x→0+ }f(x) = lim_{x→0- }f(x) = f(0)

For concavity:

Isn't the second derivative negative everywhere, except at x = 0, where it does not exist ?

- #14

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How do I know where thereis convavity, if I don't have 0s to do a number line to check where is positive or negative?

- #15

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If you can't get it, what do you have as the second derivative?

I just have to say I think this is a squirrely function...bends and corners... icky.

- #16

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I do remember not being able to solve for x.

Can someone verify for me that there are no 0s for the second derivative test?

- #17

SammyS

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There are no zeros for the second derivative, but there is a critical point.

I do remember not being able to solve for x.

Can someone verify for me that there are no 0s for the second derivative test?

Look at the second derivative (when you get back to your working paper). It should be obvious that it is negative wherever it's defined.

- #18

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There are no zeros for the second derivative, but there is a critical point.

Look at the second derivative (when you get back to your working paper). It should be obvious that it is negative wherever it's defined.

I don't understand. How can I get a criticial point if I can't get a 0?

- #19

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- #20

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Where do I plug in values in?

- #21

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There is no point of inflection.

Correct?

- #22

SammyS

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Correct.

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