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November 29, 2014

November 29, 2014

Posted by **casey** on Thursday, November 5, 2009 at 4:32pm.

- calc -
**MathMate**, Thursday, November 5, 2009 at 6:02pmThe absolute maximum and absolute minimum of implicit equations are difficult to find because of the uncertainty of the number of local maxima and minima.

For a quadratic equation, we know that there is only one single maximum/minimum which will automatically be the global value. If a global maximum falls within the interval, the smaller value of each of the limits will give the minimum within the interval.

Thus, interval = [1,3].

f(x)= -x^2 +3x

f'(x)=-2x+3

f'(x)=0 at x=1.5, falls within [1,3]

f"(x)=-2 f'(1.5) is a maximum.

Thus the maximum is at x=1.5, or f(1.5)=2.25.

The minimum is one of the two following values (evaluated at the limits of the given interval).

f(1) or f(3).

Can you take it from here?

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