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Chapter 4 17

# Chapter 4 17 - 27 28 29 30 31 32 33 SECTION 4.2 THE MEAN...

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Unformatted text preview: 27. 28. 29. 30. 31. 32. 33. SECTION 4.2 THE MEAN VALUE THEOREM 279 We use Exercise 26 with f(x) 2 \/1+ .T. g(.1;) 2 1+ ~21-zc. and a 2 0. Notice that f(0) 2 1 2 9(0) and f’(.r) 2 Nﬁ < % 2 g’(w) form > 0. So by Exercise 26. f(b) < g(b) 2> \/1+b < 1+ ébforb > 0. Another method: Apply the Mean Value Theorem directly to either f(a:) 2 1 + 53: 2 \/1 —l— .7: or g(m) 2 1 + m on [0. b]. f satisﬁes the conditions for the Mean Value Theorem. so we use this theorem on the interval [21). b]: M — f’(c) for some c 6 (—b b) But since f is odd f(2b) 2 2f(b) Substitutin‘y this into the above I) i (*5) _ . . . . . . — . 0 equation. we get W132 2 f’(c) 2> \$ 2 f’(c). Let ﬂat) 2 sins: and let I) < a. Then f(a:) is continuous on [b, a] and differentiable on (b. a). By the Mean Value Theorem. there is a number 0 E (b. a.) with sina 2 sinb 2 f(a) — f(b) 2 f’(c)(a — b) 2 (cos c)(a — (7). Thus. lsina 2 sinb| S lcoscl |b — al 3 |a — bl. Ifa < b. then |sina — sinb] 2 lsinb — sinal S [b — a] 2 Ia — b]. If a 2 b. both sides of the inequality are 0. Suppose that f’(\$) 2 0. Let 9(37) 2 at, so g'(\$) 2 (2. Then. by Corollary 7. f(:c) 2 9(m) + d, where dis a constant. so f(a:) 2 car —l— d. Fora: > 0. f(\$) 2 g(:r). so f’(m) 2 g’(\$). Form < 0. f’(:1:) 2 (l/cc)’ 2 21/m2 and g'(m) 2 (1+ 1/w)’ 2 —1/£C2, so again f/(w) 2 g’(a:). However. the domain ofg(:£) is not an interval [it is (—00. 0) U (0. 00)] so we cannot conclude that f 2 g is constant (in fact it is not). Let f(\$) 2 2sin‘1 x 2 cos’1 (1 — 2562). Then 2 4a: 2 4m INC”) ‘ — 0 since at > 0 . Thus. ’ 2 Ofo ﬁ—\$2 \/12(1—2;c2)2 \/1—\$2 2\$v1—m2 ( T ) f(3?) r all a: E (0.1). Thus. f(:z:) 2 C on (0.1). To ﬁnd C. let a: 2 0.5. Thus. 2sin’1(0.5) 2 cos—1(0.5) 2 2(%) — g 2 0 2 C. We conclude that f(a:) 2 0 force in (0. 1). By continuity of f. f(:c) 2 0 on [0,1]. Therefore. we see that f(m) 2 2 sin‘1 30 2 cos-1(12 2m2) 2 O 2> 2 sin‘1 x 2 cosT1(1 2 23:2). ‘ 21 Let f(:c) 2 arcsln<:+1) 2 2 arctanﬂJr g. Note that the domain off is [0. 00). Thus. , 1 (m+l)2(ac21) 2 1 1 1 f(:c)2 - 2x—x20.Th 1*(m21)2 (m+1)2 1+3: 2% ﬁ(a:+1) ﬁ(z—l—1) en x+1 f(\$) 2 Con (0. 00) by Theorem 5. By continuity of f, f(ac) 2 C 0n [0. 00). To ﬁnd C. we let a: 2 0 2> arcsin(—1) 2 2arctan(0) + g 2 C 2> 2% 2 0+ g 2 0 2 C. Thus. f(a:) 2 0 2> . 3:21 W arcs1n(m+1> 22arcta11\/—— 5. ...
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