`f(x) = x - x^(1/3), [0,1]` Determine whether Rolle's Theorem can be applied to `f` on the interval and, if so, find all values of `c` in the open interval `(a,b)` such that `f'(c) = 0`.

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Chapter 3, 3.2 - Problem 24 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to notice that the given function is continuous on [0,1] and differentiable on (0,1), since it is a  polynomial function.

You need to verify if f(0)=f(1), hence, you need to evaluate the values of function at x = 0 and x = 1.

`f(0) = 0 - 0^(1/3) = 0`

`f(1) = 1 - 1^(1/3) = 1-1 = 0`

Since f(0)=f(1) = 0 and the function is continuous and differentiable on the given interval, the Rolle's theorem may be applied, hence, there is a point c in (0,1), such that:

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

You need to find the derivative of the function:

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

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

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

`1 - 1/(2*root(3)(c^2)) = 0 => 1/(2*root(3)(c^2)) = 1 => (2*root(3)(c^2)) = 1 => root(3)(c^2) = 1/2`

Raise to 3rd power both sides:

`c^2 = (1/2)^3 => c^2 = 1/8 => c_1 = sqrt 1/(2sqrt2); c_2 = -1/(2sqrt2)`

Notice that `c_2 = -1/(2sqrt2)` does not belong to (0,1).

Hence, applying Rolle's theorem to the given function yields that there is `c = sqrt 1/(2sqrt2) in (0,1)` , such that `f'(c) = 0.`

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