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?

Answer 1

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].

Answer 2

mean value theorem of x^3+x-6 with range of [0,2]