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Pre-Calc Homework Solutions 223

# Pre-Calc Homework Solutions 223 - Section 5.4 1 2 1 1...

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(b) The vertex is at x 5 } 2 2( ( 2 2 1 1 ) ) } 5 2 } 1 2 } . 1 Recall that the vertex of a parabola y 5 ax 2 1 bx 1 c is at x 5 2 } 2 b a } . 2 y 1 2 } 1 2 } 2 5 } 2 4 5 } , so the height is } 2 4 5 } . (c) The base is 2 2 ( 2 3) 5 5. } 2 3 } (base)(height) 5 } 2 3 } (5) 1 } 2 4 5 } 2 5 } 12 6 5 } 53. (a) H (0) 5 E 0 0 f ( t ) dt 5 0 (b) H 9 ( x ) 5 } d d x } 1 E x 0 f ( t ) dt 2 5 f ( x ) H 9 ( x ) . 0 when f ( x ) . 0. H is increasing on [0, 6]. (c) H is concave up on the open interval where H 0 ( x ) 5 f 9 ( x ) . 0. f 9 ( x ) . 0 when 9 , x # 12. H is concave up on (9, 12). (d) H (12) 5 E 12 0 f ( t ) dt . 0 because there is more area above the x -axis than below the x -axis. H (12) is positive. (e) H 9 ( x ) 5 f ( x ) 5 0 at x 5 6 and x 5 12. Since H 9 ( x ) 5 f ( x ) . 0 on [0, 6), the values of H are increasing to the left of x 5 6, and since H 9 ( x ) 5 f ( x ) , 0 on (6, 12], the values of H are decreasing to the right of x 5 6. H achieves its maximum value at x 5 6. (f) H ( x ) . 0 on (0, 12]. Since H (0) 5 0, H achieves its minimum value at x 5 0. 54. (a) s 9 ( t ) 5 f ( t ). The velocity at t 5 5 is f (5) 5 2 units/sec.
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