- SOLUTION FOR THOMAS CALCULUS 11TH EDITION CHAPTER 1 9 HOW TO
- SOLUTION FOR THOMAS CALCULUS 11TH EDITION CHAPTER 1 9 MANUAL PDF
This fact is important because it means that for a given function f, f, if there exists a function F F such that F ′ ( x ) = f ( x ) F ′ ( x ) = f ( x ) then, the only other functions that have a derivative equal to f f are F ( x ) + C F ( x ) + C for some constant C.
SOLUTION FOR THOMAS CALCULUS 11TH EDITION CHAPTER 1 9 HOW TO
We make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum and minimum values of the function, and how to determine the shape of the graph. Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing if the derivative is negative, then the function is decreasing ( Figure 4.29). Recall that a function f f is increasing over I I if f ( x 1 ) f ( x 2 ) f ( x ) 1 > f ( x 2 ) whenever x 1 < x 2. 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.
SOLUTION FOR THOMAS CALCULUS 11TH EDITION CHAPTER 1 9 MANUAL PDF
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