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Chapter 4 | Applications of Derivatives
Case 3: The case when there exists a point x ∈ ( a , b ) such that f ( x ) < k is analogous to case 2, with maximum replaced by minimum. □ An important point about Rolle’s theorem is that the differentiability of the function f is critical. If f is not differentiable, even at a single point, the result may not hold. For example, the function f ( x ) = | x | −1 is continuous over [−1, 1] and f (−1) =0= f (1), but f ′( c ) ≠0 for any c ∈ (−1, 1) as shown in the following figure.
Figure 4.22 Since f ( x ) = | x | −1 is not differentiable at x =0, the conditions of Rolle’s theorem are not satisfied. In fact, the conclusion does not hold here; there is no c ∈ (−1, 1) such that f ′( c ) =0.
Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points c where f ′( c ) =0.
Example 4.14 Using Rolle’s Theorem
For each of the following functions, verify that the function satisfies the criteria stated in Rolle’s theorem and find all values c in the given interval where f ′( c ) =0.
a. f ( x ) = x 2 +2 x over [−2, 0] b. f ( x ) = x 3 −4 x over [−2, 2]
Solution a. Since f is a polynomial, it is continuous and differentiable everywhere. In addition, f (−2) =0= f (0). Therefore, f satisfies the criteria of Rolle’s theorem. We conclude that there exists at least one value c ∈ (−2, 0) such that f ′( c ) =0. Since f ′( x ) =2 x +2=2( x +1), we see that f ′( c ) =2( c +1) =0 implies c =−1 as shown in the following graph.
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