Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 4

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)

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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