Sunday, February 12, 2012

What are the extremes of the function 4x-8x^2?

To determine the extremes of the function, we'll have to
calculate the critical values of the function, that are the roots of the first
derivative of f(x).


We'll differentiate the function with
respect to x, to determine the 1st derivative.


f'(x) =
(4x-8x^2)'


f'(x) = 4 -
16x


We'll cancel the first
derivative:


f'(x) = 0 <=> 4 - 16x =
0


We'll divide by 4:


1 - 4x =
0


We'll subtract 1:


-4x =
-1


x = 1/4


Since there is a
single critical value, the function has a single extreme
point.


The extreme value of the function
is:


f(1/4) = 4*(1/4) -
8/16


f(1/4) = 1 - 1/2


f(1/4) =
1/2


The extreme point of the function is
represented by the pair: (1/4 ; 1/2).

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