To illustrate the Mean Value Theorem with a specific function, let's consider f(x) = x3 - x, a = 0, b = 8. Since f is a polynomial, it is continuous and differentiable for all x, so it is certainly continuous on [0, 8] and differentiable on (0, 8). Therefore, by the Mean Value Theorem, there is a number c in (0, 8) such that f(8) f(0) = f'(c)(8-0). Now f(8) 504 f(0) = 0 , and f'(x) = 3n² - 1 , so this equation becomes 7 = f'(c)(8) = 0 (8) - 504 which gives c2 21.33 , that is, c = ± 4:6188 . But c must be in (0, 8), so c = 4.62 The following figure illustrates the calculation that the tangent line at this value of c is parallel to the secant line. 700 600 500 400 300 200 100 8 Ⓡ

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