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

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

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.

oops. we wanted f'=3, not 6. adjust your solution.

To find the values of C that satisfy the Mean Value Theorem (MVT) for the function f(x) = x^3 - x over the interval [0, 2], we need to ensure that two conditions are met:

1. The function must be continuous on the closed interval [0, 2].
2. The function must be differentiable on the open interval (0, 2).

Let's check each condition:

1. Continuity:
To check if the function is continuous on [0, 2], we need to ensure that the function is defined and has no breaks or jumps within the interval.
The function f(x) = x^3 - x is a polynomial, which means it is defined for all real numbers. Therefore, the function is continuous on the interval [0, 2].

2. Differentiability:
To check if the function is differentiable on (0, 2), we need to differentiate the function and ensure that the derivative exists for all values within the interval.
The derivative of f(x) = x^3 - x can be found by using the power rule, which states that the derivative of x^n is n*x^(n-1). Applying the power rule, we get:
f'(x) = 3x^2 - 1

Now let's check if the derivative exists for all values in the interval (0, 2).
Since f'(x) = 3x^2 - 1 is also a polynomial, it is defined for all real numbers.
Therefore, the function f(x) = x^3 - x is differentiable on the interval (0, 2).

Now, since both conditions for the MVT are satisfied, we can proceed to find the values of C that satisfy the MVT.

According to the Mean Value Theorem:
If a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in the open interval (a, b) such that f'(c) = (f(b) - f(a))/(b - a).

In our case, the closed interval is [0, 2], and the open interval is (0, 2).
So, we need to find the value(s) of c in the interval (0, 2) that satisfy the equation f'(c) = (f(2) - f(0))/(2 - 0).

Calculating the values:
f(2) = (2)^3 - 2 = 8 - 2 = 6
f(0) = (0)^3 - 0 = 0 - 0 = 0

Therefore, we have:
(f(2) - f(0))/(2 - 0) = (6 - 0)/(2 - 0) = 6/2 = 3

Now we need to find the value(s) of c that satisfy f'(c) = 3.
To find these value(s), we need to solve the equation 3x^2 - 1 = 3.

3x^2 - 1 = 3
3x^2 = 4
x^2 = 4/3
x = ±√(4/3)

So, the values of C that satisfy the MVT for the function f(x) = x^3 - x over the interval [0, 2] are C = √(4/3) and C = -√(4/3).