`f(x) = sqrt(((2 - x^(2/3))^3), [-1,1]` Explain why Rolle’s Theorem does not apply to the function even though there exist `a` and `b` such that `f(a) = f(b)`.

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Chapter 3, 3.2 - Problem 4 - 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 [-1,1] and differentiable on (-1,1), since it is a  polynomial function.

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

`f(-1) = sqrt(2 - root(3)((-1)^2))^3 = 1`

`f(1) = sqrt(2 - root(3)((1)^2))^3 = 1`

Since f(-1)=f(1) = 1 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 (-1,1)` , such that:

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

You need to find the derivative of the function, using chain rule:

`f'(c) = (sqrt(2 - root(3)(c^2))^3)`

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

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

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

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

Replacing the found values in equation 2f'(c) = 0 yields:

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

Raise to 2rd power both sides:

`(2 - c^(2/3)) = 0 => c^(2/3) = 2 => c = 2^(3/2) => c = 2sqrt2`

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

Hence, applying Rolle's theorem to the given function yields that there is no values of `c in(-1,1), ` such that `f'(c) = 0` .

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