The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Therefore, f ( x ) = g ( x ) + C f ( x ) = g ( x ) + C for all x ∈ I. By Corollary 1, there is a constant C C such that h ( x ) = C h ( x ) = C for all x ∈ I. Then, h ′ ( x ) = f ′ ( x ) − g ′ ( x ) = 0 h ′ ( x ) = f ′ ( x ) − g ′ ( x ) = 0 for all x ∈ I. If f f and g g are differentiable over an interval I I and f ′ ( x ) = g ′ ( x ) f ′ ( x ) = g ′ ( x ) for all x ∈ I, x ∈ I, then f ( x ) = g ( x ) + C f ( x ) = g ( x ) + C for some constant C. The Mean Value Theorem states that if f f is continuous over the closed interval and differentiable over the open interval ( a, b ), ( a, b ), then there exists a point c ∈ ( a, b ) c ∈ ( a, b ) such that the tangent line to the graph of f f at c c is parallel to the secant line connecting ( a, f ( a ) ) ( a, f ( a ) ) and ( b, f ( b ) ). Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem ( Figure 4.25).

The Mean Value Theorem generalizes Rolle’s theorem by considering functions that do not necessarily have equal value at the endpoints. In Rolle’s theorem, we consider differentiable functions f f defined on a closed interval with f ( a ) = f ( b ) f ( a ) = f ( b ). Rolle’s theorem is a special case of the Mean Value Theorem. Find all points c c guaranteed by Rolle’s theorem. Verify that the function f ( x ) = 2 x 2 − 8 x + 6 f ( x ) = 2 x 2 − 8 x + 6 defined over the interval satisfies the conditions of Rolle’s theorem. For example, the function f ( x ) = | x | − 1 f ( x ) = | x | − 1 is continuous over and f ( −1 ) = 0 = f ( 1 ), f ( −1 ) = 0 = f ( 1 ), but f ′ ( c ) ≠ 0 f ′ ( c ) ≠ 0 for any c ∈ ( −1, 1 ) c ∈ ( −1, 1 ) as shown in the following figure. If f f is not differentiable, even at a single point, the result may not hold. Ĭase 3: The case when there exists a point x ∈ ( a, b ) x ∈ ( a, b ) such that f ( x ) < k f ( x ) < k is analogous to case 2, with maximum replaced by minimum.Īn important point about Rolle’s theorem is that the differentiability of the function f f is critical. Because f f has a maximum at an interior point c, c, and f f is differentiable at c, c, by Fermat’s theorem, f ′ ( c ) = 0. As a result, the absolute maximum must occur at an interior point c ∈ ( a, b ). Therefore, the absolute maximum does not occur at either endpoint. There exists x ∈ ( a, b ) x ∈ ( a, b ) such that f ( x ) k, f ( x ) > k, the absolute maximum is greater than k.There exists x ∈ ( a, b ) x ∈ ( a, b ) such that f ( x ) > k.f ( x ) = k f ( x ) = k for all x ∈ ( a, b ).There then exists at least one c ∈ ( a, b ) c ∈ ( a, b ) such that f ′ ( c ) = 0. Let f f be a continuous function over the closed interval and differentiable over the open interval ( a, b ) ( a, b ) such that f ( a ) = f ( b ).

Ishowu instant 1.1.1 meaning and uses