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.
x= +/- 2/sqrt3
-2/sqrt3 is outside of interval so just use 2/sqrt3 (I think that's considered a critical point)
f(0) = 5
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
Function has local maximum for x=1.1547005383792515290182975610039
If you want to see graphs of function in google type:
functios graphs online
When you see list of results click on:
When page be vopen in blue rectangle type:
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
Function has local minimum for x=1.1547005383792515290182975610039
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?
Calculus - Janet, Monday, April 18, 2011 at 1:34pm
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?
Calculus - Bosnian, Wednesday, April 20, 2011 at 2:52am
If you try to see graph of that function you can see that function:
havent absolute maximum or absolute minimum.
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