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max and min of a function

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 ??
0
c_hockland
Asked:
c_hockland
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1 Solution
 
ozoCommented:
f(0)=0
f(2)=4
f(4)=0
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c_hocklandAuthor Commented:
so , which is min and which is max?

4 is max. and 0 = min , right ??
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sunnycoderCommented:
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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ozoCommented:
maximum is the highest value
minumum is the lowest value
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sunnycoderCommented:
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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