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=

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- Dec 12th 2008, 09:54 AMbgwarhawkFind 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, 10:11 AMmoolimanj
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, 10:15 AMbgwarhawk
- Dec 12th 2008, 10:34 AMchabmgph
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, 10:39 AMbgwarhawk
- Dec 12th 2008, 10:46 AMchabmgph