Posted by Sam on Thursday, March 17, 2011 at 2:41am.
find the derivative
dy/dx = 4x^3 - 30x^2 + 36x
set that equal to zero
4x^3 - 30x^2 + 36x= 0
2x(2x^2 - 15x + 18) = 0
2x(x-6)(2x-3) = 0
x = 0 or x = 6 or x = 3/2 , but x=0 lies outside our interval
so f(6) = ....
f(3/2) = ...
and for the end-values
f(1) = ...
f(8) = ...
you do the arithmetic, and decide which is the highest and lowest f(x) value.
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