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March 30, 2015

March 30, 2015

Posted by **Sam** on Thursday, March 17, 2011 at 2:41am.

- Calculus -
**Reiny**, Thursday, March 17, 2011 at 9:36amfind 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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