Posted by **Daniella** on Tuesday, November 13, 2012 at 11:42pm.

Consider f(x)=x^3-x over the interval [0,2].

Find all the values of C that satisfy the Mean Value Theorem (MVT)

- Calculus -
**Steve**, Wednesday, November 14, 2012 at 10:22am
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.

- Calculus -
**Steve**, Wednesday, November 14, 2012 at 10:23am
oops. we wanted f'=3, not 6. adjust your solution.

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