chapter 2 62

# chapter 2 62 - 126 CHAPTER 2 LIMITS AND DERIVATIVES 36...

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Unformatted text preview: 126 CHAPTER 2 LIMITS AND DERIVATIVES 36. Since f(a:) : :32 sin(1/ac) when x ¢ 0 and f(0) = 0. we have , _ , f(0+h)—f(0) . hzsin(1/h)—0 . , . 1 f (0) —}lL111() ——f;—-— : A1310 —T— : Agbh3111(1/lt). Since —1 S sin h S 1. we have . 1 . 1 — |h| S |h| sm h g |h| : — |h| g hsm 71 g |h|. Because ’lLini)(— |h|) = 0 and lint) |h| = 0. we know that . . 1 I Ill—IR) (h Sin h) : 0 by the Squeeze Theorem. Thus. f (0) = 0. 2.9 The Derivative as a Function 1. Note: Your answers may vary depending on your estimates. By estimating the slopes of tangent lines on the graph of f. it appears that (a)f’(1)z —2 (b) f’(2) w 0-8 (c) f’(3) ~ —1 (d) f’(4) z —05 2. Note: Your answers may vary depending on your estimates. By estimating the slopes of tangent lines on the graph of f. it appears that (a) f’(0) m *3 (b) f’(1) % 0 (c) f’(2) % 15 (d) f’(3) % 2 (e) f’(4) m 0 (f) f’(5) z —1.2 3. It appears that f is an odd function. so f’ will be an even function—that is. f’(—a) : f'(0«)- (a) f’t—S) ~ 1-5 (b) f’(—2) w 1 (c) f’(—1) m 0 (d) f’(0) % #4 (e) f’(1) % 0 (f) f’(2) a: 1 (g) f'(3) % 1-5 4. (a)' : 11. since from left to right. the slopes of the tangents to graph (a) start out negative. become 0. then positive. then 0. then negative again. The actual function values in graph 11 follow the same pattern. (b)’ : IV. since from left to right. the slopes of the tangents to graph (b) start out at a ﬁxed positive quantity, then suddenly become negative. then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents. (c)’ : 1. since the slopes of the tangents to graph (c) are negative for 95 < 0 and positive for x > 0. as are the function values of graph 1. (d)’ : 111. since from left to right. the slopes of the tangents to graph ((1) are positive. then 0, then negative. then 0. then positive. then 0. then negative again. and the function values in graph III follow the same pattern. ...
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