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# Business Calc Homework w answers_Part_13 - 61 Section 2.4...

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11. (a) lim h 0 } y (2 1 h h ) 2 y (2) } 5 lim h 0 5 lim h 0 5 lim h 0 } 1 h 2 ( h ( h 1 1 1) 1) } 5 lim h 0 1 2 } h 1 1 1 } 2 5 2 1 (b) The tangent line has slope 2 1 and passes through (2, y (2)) 5 (2, 1). y 5 2 ( x 2 2) 1 1 y 5 2 x 1 3 (c) The normal line has slope 2 } 2 1 1 } 5 1 and passes through (2, y (2)) 5 (2, 1). y 5 1( x 2 2) 1 1 y 5 x 2 1 (d) [ 2 4.7, 4.7] by [ 2 3.1, 3.1] 12. (a) lim h →0 } y (0 1 h h ) 2 y (0) } 5 lim h →0 5 lim h →0 } h 2 2 h 3 h } 5 lim h 0 ( h 2 3) 5 2 3 (b) The tangent line has slope 2 3 and passes through (0, y (0)) 5 (0, 2 1). y 5 2 3( x 2 0) 2 1 y 5 2 3 x 2 1 (c) The normal line has slope 2 } 2 1 3 } 5 } 1 3 } and passes through (0, y (0)) 5 (0, 2 1). y 5 } 1 3 } ( x 2 0) 2 1 y 5 } 1 3 } x 2 1 (d) [ 2 6, 6] by [ 2 5, 3] 13. (a) Near x 5 2, f ( x ) 5 ) x ) 5 x . lim h 0 } f (2 1 h h ) 2 f (2) } 5 lim h 0 } (2 1 h h ) 2 2 } 5 lim h 0 1 5 1 (b) Near x 5 2 3, f ( x ) 5 ) x ) 5 2 x . lim h 0 5 lim h 0 } (3 2 h h ) 2 3 } 5 lim h 0 2 1 5 2 1 14. Near x 5 1, f ( x ) 5 ) x 2 2 ) 5 2 ( x 2 2) 5 2 2 x . lim h 0 } f (1 1 h h ) 2 f (1) } 5 lim h 0 5 lim h 0 } 1 2 h h 2 1 } 5 lim h 0 2 1 5 2 1 15. First, note that f (0) 5 2. lim h 0 2 } f (0 1 h h ) 2 f (0) } 5 lim h 0 2 } (2 2 2 h 2 h h 2 ) 2 2 } 5 lim h 0 2 } 2 2 h h 2 h 2 } 5 lim h 0 2 ( 2 2 2 h ) 5 2 2 lim h 0 1 } f (0 1 h h ) 2 f (0) } 5 lim h 0 1 } (2 h 1 h 2) 2 2 } 5 lim h 0 1 2 5 2 No, the slope from the left is 2 2 and the slope from the right is 2. The two-sided limit of the difference quotient does not exist. 16. First, note that f (0) 5 0. lim h 0 2 } f (0 1 h h ) 2 f (0) } 5 lim h 0 2 } 2 h h 2 0 } 5 2 1 lim h 0 1 } f (0 1 h h ) 2 f (0) } 5 lim h 0 1 } ( h 2 2 h h ) 2 0 } 5 lim h 0 1 ( h 2 1) 5 2 1 Yes. The slope is 2 1. 17. First, note that f (2) 5 } 1 2 } lim h 0 2 } f (2 1 h h ) 2 f (2) } 5 lim h 0 2 5 lim h 0 2 } 2 2 2 h (2 (2 1 1 h h ) ) } 5 lim h 0 2 } 2 h ( 2 2 1 h h ) } 5 lim h 0 2 2 } 2(2 1 1 h ) } = 2 } 1 4 } } 2 1 1 h } 2 } 1 2 } }} h [2 2 (1 1 h )] 2 (2 2 1) }}} h f ( 2 3 1 h ) 2 f ( 2 3) }}} h ( h 2 2 3 h 2 1) 2 ( 2 1) }}} h } h 1 1 1 } 2 1 }} h } (2 1 1 h ) 2 1 } 2 } 2 2 1 1 } }}} h Section 2.4 61

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17. continued lim h 0 1 } f (2 1 h h ) 2 f (2) } 5 lim h 0 1 5 lim h 0 1 } [4 2 (2 4 1 h h )] 2 2 } 5 lim h 0 1 } 2 4 h h } 52 } 1 4 } Yes. The slope is 2 } 1 4 } . 18. No. The function is discontinuous at x 5 } 3 4 p } because lim x (3 p /4) 2 f ( x ) 5 lim x (3 p /4) 2 sin x 5 sin } 3 4 p } 5 } ˇ 2 2 w } but f 1 } 3 4 p } 2 5 cos } 3 4 p } 5 2 } ˇ 2 2 w } . 19. (a) lim h 0 } f ( a 1 h h ) 2 f ( a ) } 5 lim h 0 5 lim h 0 5 lim h 0 } 2 ah 1 h h 2 } 5 lim h 0 (2 a 1 h ) 5 2 a (b) The slope of the tangent steadily increases as a increases.
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Business Calc Homework w answers_Part_13 - 61 Section 2.4...

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