`f(x) = x^(2/3), [0,1]` Determine whether the Mean Value Theorem can be applied to `f` on the closed interval `[a,b]`. If the Mean Value Theorem can be applied, find all values of `c` in the...

`f(x) = x^(2/3), [0,1]` Determine whether the Mean Value Theorem can be applied to `f` on the closed interval `[a,b]`. If the Mean Value Theorem can be applied, find all values of `c` in the open interval `(a,b)` such that `f'(c) = (f(b) - f(a))/(b - a)`. If the Mean Value Theorem cannot be applied, explain why not.

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Textbook Question

Chapter 3, 3.2 - Problem 41 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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The mean value theorem is applicable to the given function, since it is a polynomial function. All polynomial functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval.

The mean value theorem states:

`f(b) - f(a) = f'(c)(b-a)`

Replacing 1 for b and 0 for a, yields:

`f(1) - f(0) = f'(c)(1-0)`

Evaluating f(1) and f(0) yields:

`f(1) = 1^(2/3) => f(1) =1`

`f(0) = 0`

You need to evaluate f'(c):

`f'(c) = (c^(2/3))' => f'(c) = (2/3)c^(2/3 - 1) => f'(c) = (2/3)*(c^(-1/3)) => f'(c) = 2/(3 root(3) c)`

Replacing the found values in equation `f(1) - f(0) = f'(c)(1-0):`

`1 - 0 = (2/(3 root(3) c))(1-0) => 1 = 2/(3 root(3) c) =>3 root(3) c = 2 => root(3) c = 2/3 => c = (2/3)^3 => c = 8/27 in [0,1]`

Hence, in this case, the mean value theorem can be applied and the value of c is `c = 8/27` .

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