# Using the principle of the mean-value theorem on the indicated interval, how do you find all numbers c that satisfy the conclusion of the theorem #f(x)=x^3-2x^2#; [0, 2]?

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To find all numbers (c) that satisfy the conclusion of the Mean Value Theorem for the function (f(x) = x^3 - 2x^2) on the interval ([0, 2]), we first need to check if the function satisfies the conditions of the Mean Value Theorem on the interval ([0, 2]). The function (f(x) = x^3 - 2x^2) is continuous on ([0, 2]) because it is a polynomial function, and it is differentiable on ((0, 2)) because it is a differentiable function.

Next, we find the average rate of change of the function over the interval ([0, 2]) using the formula for the Mean Value Theorem:

[f'(c) = \frac{f(2) - f(0)}{2 - 0}]

Now, we find the derivative of (f(x)) and evaluate it at (x = c):

[f'(x) = 3x^2 - 4x]

Now, we evaluate (f'(c)) and (f(2) - f(0)):

[f'(c) = 3c^2 - 4c] [f(2) - f(0) = (2)^3 - 2(2)^2 - (0)^3 + 2(0)^2] [f(2) - f(0) = 8 - 8 - 0 + 0] [f(2) - f(0) = 0]

Now, we set the two expressions equal to each other and solve for (c):

[3c^2 - 4c = 0] [c(3c - 4) = 0]

So, either (c = 0) or (3c - 4 = 0). Solving the second equation, we get:

[3c - 4 = 0] [3c = 4] [c = \frac{4}{3}]

Thus, the numbers (c) that satisfy the conclusion of the Mean Value Theorem for the function (f(x) = x^3 - 2x^2) on the interval ([0, 2]) are (c = 0) and (c = \frac{4}{3}).

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