You first show that it satisfies the three requirements: f(a) = f(b), f is continuous on [a,b], and f is differentiable on (a,b).

As f(x) = x^3 - x^2 -2x +8, f(x) is a polynomial, so it is continuous and differentiable everywhere.

f(0) = 8

f(2) = 2^3 - 2^2 - 2*2 + 8 = 8 - 4 - 4 + 8 = 8

Thus, as f(0) = 8 and f(2) = 8, f(0) = f(2)

Michael S, was correct about there being one critical value in the interval.

From the quadratic formula, the zeroes are at (1 +- sqrt(7))/3, and the value in the interval is at (1 + sqrt(7))/3, which is approximately 1.22