Pre-Calc Homework Solutions 266

# Pre-Calc Homework Solutions 266 - 9. } x x 2 2 2 1 4 2 x }...

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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).
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Unformatted text preview: 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 5 2 12, we have A 5 3 and B 5 2 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...
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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