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]?

Answer 1
The conclusion of the Mean Value Theorem for #f# on #[a,b]# says:
there is a #c# in #(a,b)# such that #f'(c) = (f(b)-f(a))/(b-a)#
(The theorem make no guarantees about our ability to find the value(s) of #c#)
Find the value(s) of #c# does not use the Mean Value Theorem, it uses the derivative and some algebra.
For #f(x) = x^3-2x^2# on #[0,2]#, the conclusion of MVT says:
there is a #c# in #(0,2)# such that #f'(c) = (f(2)-f(0))/(2-0)#
To find the #c# (or #c#'s) find #f'(x)#, do the arithmetic on the right and solve the resulting equation. In this case, solve
#3x^2-4x = 0#
The solutions are #0# and #4/3#.
The #c# mentioned in MVT must be in #(0,2)#, so #4/3# is the only value of #c#
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Answer 2

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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