Posted by **Janet** on Sunday, April 17, 2011 at 11:29pm.

Find any absolute maximum and minimum and local maximum and minimum for the function f(x)=x^3-4x+5 on the interval [0,5). Make sure to prove that these points are max/min values.

THis is what I did.

f'(x)=3x^2-4=0

x= +/- 2/sqrt3

-2/sqrt3 is outside of interval so just use 2/sqrt3 (I think that's considered a critical point)

f(2/sqrt3)= 1.921

f(0) = 5

f(5)=110

I'm not sure what to do with all these numbers. I think since 1.921 is the smallest of the numbers that 2/sqrt 3 is the minimum, but is it the local or absolute.

It looks like 5 is the max, but again local or absolute and can I really include this to be max since it isn't even included in the interval [0,5)

Can someone give me some guidance or show me some steps to solve this.

- Calculus -
**Bosnian**, Monday, April 18, 2011 at 7:26am
Functionn has local maximum or minimum where first derivation=0

If second derivate<0 that is local maximum

If second derivate>0 that is local minimum

First derivation=3x^2-4=0

x=+/- 2/sqroot(3)

sqroot(3)=1.7320508075688772935274463415059

x=+/- 2/1.7320508075688772935274463415059=

1.1547005383792515290182975610039

Second derivate=6x

6*1.1547005383792515290182975610039=

6.9282032302755091741097853660235>0

Function has local maximum for x=1.1547005383792515290182975610039

f(x)max=f(1.1547005383792515290182975610039)=

1.9207985643219959226178731706562

If you want to see graphs of function in google type:

functios graphs online

When you see list of results click on:

rechneronline.de/function-graphs/

When page be vopen in blue rectangle type:

x^2x^3-4x+5

In Range x-axis from type -4 to 6

In Range y-axis from type -1 to 9

Then click option Draw

- Calculus -
**Bosnian**, Monday, April 18, 2011 at 7:31am
Sorry:

Second derivate=6x

6*1.1547005383792515290182975610039=

6.9282032302755091741097853660235>0

Function has local minimum for x=1.1547005383792515290182975610039

f(x)min=f(1.1547005383792515290182975610039)=

1.9207985643219959226178731706562

- Calculus -
**Anonymous**, Monday, April 18, 2011 at 10:57am
So the local min and max are both 2/sqrt3. what about the absolite values or do the not exists?

Thanks.

- Calculus -
**Janet**, Monday, April 18, 2011 at 1:34pm
Bosnian,

Thanks for your help. I just want to make sure I understand. The local min and max are the same f(2/sqrt3) = 1.92

Are there any absolute maximum or minimum?

Thanks again.

- Calculus -
**Bosnian**, Wednesday, April 20, 2011 at 2:52am
If you try to see graph of that function you can see that function:

x^3-4x+5

havent absolute maximum or absolute minimum.

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