f(0) = 0
f(2) = 6
the slope of the secant is thus (6-0)/(2-0) = 3
f'(x) = 3x^2 - 1
The only place in [0,2] where f'=0 is at x = 1/√3
f' > 0 only on [1/√3,2]
f' = 6 when 3x^2-1=6, or x = √(7/3) = 1.527
That is the only C in [0,2] which satisfies the MVT.
Consider f(x)=x^3-x over the interval [0,2].
Find all the values of C that satisfy the Mean Value Theorem (MVT)
2 answers
oops. we wanted f'=3, not 6. adjust your solution.