EXAMPLE 3 To illustrate the Mean Value Theorem with a specific function, let's consider f(x) = x³-x, a=0, b 5. Since f is a polynomial, it is continuous and differentiable for all x, so it is certainly continuous on [0, 5] and differentiable on (0, 5). Therefore, by the Mean Value Theorem, there is a number c in (0,5) such that = f(5) f(0) f'(c)(5-0). - Now f(5) = ' f(0) - and f'(x) = ' so this equation becomes = f'(c)(5)= which gives c²= C= that is, c± But c must be in (0, 5), so The figure illustrates this calculation: The tangent line at this value of c is parallel to the secant line. EXAMPLE 5 Suppose that f(0) -8 and f'(x) S9 for all values of x. How large can (5) possibly be? SOLUTION We are given that f is differentiable (and therefore continuous) everywhere. In particular, we can apply the Mean Value Theorem on the interval [0, 5]. There exists a number c such that f(5) - f(0) = f(c)( - So f(5) = f(0) + f'(c) = -8 + f'(c). We are given that f'(x) s 9 for all x, so in particular we know that f'(c) s this inequality by 5, we have (5)=-8+ 5f'(c) s f(c) s-8+ The largest possible value for f(5) is So Multiplying both sides of