Find the absolute maximum and absolute minimum

• Dec 12th 2008, 08:54 AM
bgwarhawk
Find the absolute maximum and absolute minimum
f(x)= 2x^4 - 16x^2 +3; on 0,2

I got to 8x^3 -32x=0 what do I do now?

absolute maximum=
absolute minimum=
• Dec 12th 2008, 09:11 AM
moolimanj
Use the maximum-minimum principle:

calculate f(x)'

calculate a value c in the interval [0,2] such that f'(0)=c or doesnt exist

list endpoints of interval: 0,c1,c2,c3....,2

express f(x) for each value of c above

The largest is absolute maximum and the smallest is absolute minimum

OR you could just differentiate again OR just draw a graph to check your answer. Have you got mathcad or a similar graphing programme? There are online sites that calculate the graph based on an inputted function (but dont know if I can mention here - just google)
• Dec 12th 2008, 09:15 AM
bgwarhawk
Quote:

Originally Posted by moolimanj
Use the maximum-minimum principle:

calculate f(x)'

calculate a value c in the interval [0,2] such that f'(0)=c or doesnt exist

list endpoints of interval: 0,c1,c2,c3....,2

express f(x) for each value of c above

The largest is absolute maximum and the smallest is absolute minimum

OR you could just differentiate again OR just draw a graph to check your answer. Have you got mathcad or a similar graphing programme? There are online sites that calculate the graph based on an inputted function (but dont know if I can mention here - just google)

What? I just need f'(x)
• Dec 12th 2008, 09:34 AM
chabmgph
Quote:

Originally Posted by bgwarhawk
f(x)= 2x^4 - 16x^2 +3; on 0,2

I got to 8x^3 -32x=0 what do I do now?

absolute maximum=
absolute minimum=

Quote:

Originally Posted by bgwarhawk
What? I just need f'(x)

f'(x)=8x^3 -32x like you had it.

Set f'(x)=0 and solve for x. That will give you the x value(s) where your local max/min occurs.

But since f(x) is on an closed interval (I believe you meant to write [0,2]?), you also need to check the endpoints too.
• Dec 12th 2008, 09:39 AM
bgwarhawk
Quote:

Originally Posted by chabmgph
f'(x)=8x^3 -32x like you had it.

Set f'(x)=0 and solve for x. That will give you the x value(s) where your local max/min occurs.

But since f(x) is on an closed interval (I believe you meant to write [0,2]?), you also need to check the endpoints too.

What do I do with the cube root of 4x?
• Dec 12th 2008, 09:46 AM
chabmgph
Quote:

Originally Posted by bgwarhawk
What do I do with the cube root of 4x?

$8x^3-32x=0$
$8x(x^2-4)=0$
$8x=0$, $x^2-4=0$
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