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SCAN0028 - Section 2.8 Problems fiom page 146 3 3 3 2 2 3...

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Unformatted text preview: Section 2.8 Problems fiom page 146 3 3 3 2 2 3 3 8. y' 11.1112(x h) 5(x h) (2x 5x) limzx 6xh 6xh 2h 5x 5h 2x 5x hot) h hoO h 2 2 y, Emma): 6x11 2h 5) hot) h 6362 5 y'( 1) 1, so the equation of the tangent line is y x 4 2(x h) 2x 2(x h)(x 1)2 2x(x h 1)2 10 , lim(x h 1)2 (x 1)2 lim (x h 1)2(x 1)2 i y 1100 h hoO h 2x3 4x2 2x 2th 4xh 2h 2x3 4th 2th 4xh 2x (x h 1)2(x 1)2 y‘ m———————————-——-———h 2 2 ‘ y' lim 2x h 2h ljm 2(x 1) lim 2(x 1)(x 1) 2(x 1) h°0h(x h 1)2(x 1)2 h°0h(x h 1)2(x 1f hooh(x h l)2(x 1)2 (x n3 y'(O) 2 , so the equation of the tangent line is y 2x 32. at x = 0 (discontinuity) at x = 3 (vertical tangent) 34. at x = — 1 (discontinuity) at x = 2 (sham corner) 1 l x (x h) 42, f'(x) lim—————x h x uni—"(x h) h'm—————h lim 1 i2 ho 0 h ho 0 [1 ho 0 xh<x h) 110 0 x(x h) x 1 1 x2 (x h)2 2 —2 2 2 f"(x) h'm (x h) x lim x (x [2) lim h(2x h)2 lim 223: h _23_ ho o h ho 0 h ho 0x2h(x [1) ho o x (x h)2 x PDF created with pdfFactory trial version wwgdffactomcom ...
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