This preview shows pages 1–3. Sign up to view the full content.

6. Use NDER f ( x ), or calculate the derivative as follows. f 9 ( x ) 5 } d d x } } 1 1 5 5 e 0 2 0.1 x } 5 5 } (1 1 25 5 e 2 e 2 0. 0 1 . x 1 x ) 2 } [ 2 30, 70] by [ 2 0.5, 2] (a) ( 2‘ , ) (b) None 7. Use NDER(NDER f ( x )), or calculate the second derivatives as follows. f 0 ( x ) 5 } d d x } } (1 1 25 5 e 2 e 2 0. 0 1 . x 1 x ) 2 } 5 5 5 [ 2 30, 70] by [ 2 0.08, 0.08] Locate the inflection point using graphical methods, or analytically as follows. f 0 ( x ) 5 0 5 0 2.5 e 2 0.1 x (5 e 2 0.1 x 2 1) 5 0 e 2 0.1 x 5 } 1 5 } 2 0.1 x 52 ln 5 x 5 10 ln 5 < 16.094 (a) Since f 0 ( x ) . 0 for x , 10 ln 5, the graph of f is concave up on the interval ( 2‘ , 10 ln 5), or approximately ( 2‘ , 16.094). (b) Since f 0 ( x ) , 0 for x . 10 ln 5, the graph of f is concave down on the interval (10 ln 5, ), or approximately (16.094, ). 8. Using the result of the previous exercise, the inflection point occurs at x 5 10 ln 5. Since f (10 ln 5) 5 } 1 1 5 5 0 e 2 ln 5 }5 25, the point of inflection is (10 ln 5, 25), or approximately (16.094, 25). 9. } x x 2 2 2 1 4 2 x } 5 } A x } 1 } x 2 B 4 } x 2 12 5 A ( x 2 4) 1 Bx x 2 12 5 ( A 1 B ) x 2 4 A Since A 1 B 5 1 and 2 4 A 12, we have A 5 3 and B 2. 12.5 e 2 0.2 x 2 2.5 e 2 0.1 x }}} (1 1 5 e 2 0.1 x ) 3 12.5 e 2 0.2 x 2 2.5 e 2 0.1 x }}} (1 1 5 e 2 0.1 x ) 3 2 2.5 e 2 0.1 x [(1 1 5 e 2 0.1 x ) 2 2(5 e 2 0.1 x ] }}}} (1 1 5 e 2 0.1 x ) 3 (1 1 5 e 2 0.1 x ) 2 (25 e 2 0.1 x )( 2 0.1) 2 (25 e 2 0.1 x )(2)(1 1 5 e 2 0.1 x )(5 e 2 0.1 x )( 2 0.1) }}}}}}}} (1 1 5 e 2 0.1 x ) 4 (1 1 5 e 2 0.1 x )(0) 2 (50)(5 e 2 0.1 x )( 2 0.1) }}}} (1 1 5 e 2 0.1 x ) 2 266 Section 6.5

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
10. } x 2 2 x 1 1 x 1 2 6 6 }5} x 1 A 3 } 1 } x 2 B 2 } 2 x 1 16 5 A ( x 2 2) 1 B ( x 1 3) When x 52 3, the equation becomes 10 5 A , and when x 5 2, the equation becomes 20 5 5 B . Thus, A 2 and B 5 4. Section 6.5 Exercises 1. (a) } d d P t } 5 0.025 P (b) Using the Law of Exponential Change from Section 6.4, the formula is P 5 75,000 e 0.025 t (c) [0, 100] by [0, 1,000,000] 2. (a) } d d P t } 5 0.019 P (b) Using the Law of Exponential Change from Section 6.4, the formula is P 5 110,000 e 0.019 t . (c) [0, 100] by [0, 1,000,000] 3. (a) } d d P t } 5 } M k } P ( M 2 P ) } d d P t } 5 } 0 2 . 0 0 0 5 } P (200 2 P ) } d d P t } 5 0.00025 P (200 2 P ) (b) P 5 } 1 1 M Ae 2 kt } P 5 } 1 1 2 A 0 e 0 2 0.05 t } Initial condition: P (0) 5 10 10 5 } 1 1 20 A 0 e 0 } 1 1 A 5 } 2 1 0 0 0 } 5 20 A 5 19 Formula: P 5 } 1 1 1 2 9 0 e 0 2 0.05 t } (c) [0, 100] by [0, 250] 4. (a) } d d P t } 5 } m k } P ( M 2 P ) } d d P t } 5 } 0 1 .
This is the end of the preview. Sign up to access the rest of the document.