Calc
posted by Erica .
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?

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)=(1x)³, the chord between 0 and 3 has a slope of
s=(f(3)f(0))/(30)=3
The value(s) of c required must satisfy
f'(c)=3
So, differentiate with respect to x:
f'(x)=3(1x)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]. 
mean value theorem of x^3+x6 with range of [0,2]
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