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^34x+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^24=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^24=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/functiongraphs/
When page be vopen in blue rectangle type:
x^2x^34x+5
In Range xaxis from type 4 to 6
In Range yaxis 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^34x+5
havent absolute maximum or absolute minimum.
Answer This Question
Related Questions
 Calculus  Find any absolute max/min and local max/min for the function f(x)=x^3...
 Calculus (pleas help!!!)  Find the absolute maximum and absolute minimum ...
 Calculus (pleas help!!!)  Find the absolute maximum and absolute minimum values...
 Calculus  Find the absolute maximum and absolute minimum values of the function...
 Please check my Calculus  1. Find all points of inflection: f(x)=1/12x^42x^2+...
 Calculus  Consider the function g(x) = sinxcosx. a. Find an equation of the ...
 calculus  2) Let g(s)= t(4−t)^1/2 on the interval [0,2]. Find the ...
 Caluclus  Find the absolute maximum and absolute minimum of f on the interval...
 Math  I need an expiation on a math concept, because I know the right answer, I...
 calculus  a function has a local maximum at x=2 and x=6 and a local minimum at...
More Related Questions