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December 20, 2014

December 20, 2014

Posted by **Erica** on Sunday, January 23, 2011 at 12:13am.

- Mean-value theorem -
**MathMate**, Sunday, January 23, 2011 at 10:09amThe mean value theorem states that a continuous function between x=a and x=b will have at least one tangent parallel to the chord AB.

For f(x)=(1-x)³, the chord between 0 and 3 has a slope of

s=(f(3)-f(0))/(3-0)=-3

The value(s) of c required must satisfy

f'(c)=-3

So, differentiate with respect to x:

f'(x)=-3(1-x)²

and solve for

f'(x)=-3

to get

x=0 and x=2

Reject any solution that is not on the interval [0,3].

- Calc -
**Noran**, Monday, March 28, 2011 at 1:25pmmean value theorem of x^3+x-6 with range of [0,2]

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