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Chapter 4 14

# Chapter 4 14 - _i 276 CHAPTER 4 APPLICATIONS OF...

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Unformatted text preview: ________________________________________________________i 276 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION f(8)—f(0) 6—4 1 7_ ___ ! __ _ _ 8 0 8 4. lhe values ofc which satisfy f’(c) = i seem to be about c : 0.8. 3.2. 4.4. and 6.1. 9. (a). (b) The equation of the secant line is 8.5g5 y’54 8!1(;1:—1) 41> y:%\$+%. 10 / / 0 10 10. (a) 5 It seems that the tangent lines are parallel to the secant at ac m \$1.2. f(7)—f(1) 2—5 1 8- Z -— : __ ————7 _ 1 6 2. The values ofc which satisfy f’(c) = —% seem to be about c : 1.1. 2.8. 4.6. and 5.8. (c) f(m) : m+4/33 => f'(a:) : 1 — 4/302. Sof/(c)=% c:2\/2, and f(c) : 2 x/2 + ﬁi : 3 x/i Thus. an equation of the => 02:8 : tangentlineisy—3x/2: ﬂan—2M2) 4? y:%x+2\/2. (b) The slope of the secant line is 2. and its equation is 31:21:. f(m) :m3—2\$ :> f’(x) 23:32 A2. so we solve f/(c) : 2 => 362 : 4 ﬁ c : iiﬁ % 1.155. Our estimates were off by about 0.045 in each case. 11. f(a:) : 3x2 + 21: + 5. [—1. 1]. f is continuous on [—1, 1] and differentiable on (#1., 1) since polynomials are continuous and differentiable on IR. f’ (c) 6c : 0 <:> c : 0. which is in (*1. 1). : f(b)*f(a) c) 6c+2: f(1)’f(—1) Z 10#6 : bed 2 <:> 1 _ (—1) 2 ...
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