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max and min of a function

Posted on 2006-11-06
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Last Modified: 2012-06-21
f(x)=4x-x^2
i need max and min   [0,4]

f' = 4-2x

critical numbers

0,2,4

f(0)=4  max
f(2)=0
f(4)=-4   min

am i correct ??
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Question by:c_hockland
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5 Comments
 
LVL 84

Expert Comment

by:ozo
ID: 17887475
f(0)=0
f(2)=4
f(4)=0
0
 

Author Comment

by:c_hockland
ID: 17887480
so , which is min and which is max?

4 is max. and 0 = min , right ??
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LVL 45

Expert Comment

by:sunnycoder
ID: 17887535
Hi c_hockland,

f(0) = 4*0 - 0^2 = 0
f(2) = 4*2 - 2^2 = 4
f(4) = 4*4 - 4^2 = 0

4-2x =0
=> x = 2

f'(1.9) -> +ve
f'(0) -> 0
f'(2.1) -> -ve
Increases from left and decreases to right - maxima

Cheers!
sunnycoder
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LVL 84

Expert Comment

by:ozo
ID: 17887545
maximum is the highest value
minumum is the lowest value
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Accepted Solution

by:
sunnycoder earned 500 total points
ID: 17887547
you can determine if a given point is local maxima or minima using same technique ...
Here is a simple tutorial for the same
http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-max-min-feb04.pdf
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