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November 28, 2014

November 28, 2014

Posted by **Malika** on Saturday, March 3, 2012 at 12:45am.

f (x) = x^3 − 12x, [0, 4]

minimum (x, y) =( )

maximum (x, y) = ( )

I can't even give a setup cause I am lost please help!

- Math -
**David Q/R**, Saturday, March 3, 2012 at 3:26pmf(x) = x^3 - 12x. You can find the maxima and minima by differentiating it (I'm assuming here that you can use calculus) to get f'(x) = 3x^2 - 12, and solving f'(x) = 0, so 3x^2 - 12 = 0, so 3x^2 = 12, so x^2 = 4, so x is either +2 or -2. But only one of those lies in the interval [0, 4], namely x = +2. The original function has at most two turning points (because that's the order of the derivative function), and only one of them lies within the interval, so any other maxima or minina must lie at the ends of the interval, i.e. either x=0 or x=4. At x=0, f(x) = 0. At x=4, f(x) = 64 - 48 = 16. At x=2, f(x) = 8 - 24 = -16. So the minimum must be -16 at x=2, and the maximum is +16 at x=4. Done.

- Math -
**Malika**, Sunday, March 4, 2012 at 4:50pmThank you :)

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