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

Given function f defined by f(x) = ( 1- x)³. What are all values of c, in the closed interval [0,3], that satisfy the conditions of the Mean Value Theorem?

- Mean-value theorem -
**MathMate**, Sunday, January 23, 2011 at 10:09am
The 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 1:25pm
mean value theorem of x^3+x-6 with range of [0,2]

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