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chapter 1 7 - 39 41 42 45 47 SECT|0N1.1 FOUR WAYS TO...

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Unformatted text preview: 39. 41. 42. 45. 47. SECT|0N1.1 FOUR WAYS TO REPRESENTAFUNCTION 7 —1 if a: g —1 f(x)={m2+2 ifmg‘l 404(93): 3w+2 if—1<z<1 I “0‘1 7—2x ifol Note that for ac = —1, both x + 2 and m2 are Domain is R. equal to 1. Domain is R. . _ 1 . Recall that the slope m of a line between the two points ($1, yl) and ($2,112) 15 m = :1: _ :1 and an equatlon of . . . —6 — 1 7 the line connecting those two points is y — y1 = m(m — 271). The slope of this 11ne segment 18 m = _6’ so an equation is 3; ~ 1 = —%(m + 2). The function is f(m) = —%m — 1;, —2 3 an S 4. — —2 5 . . . . The slope of this line segment is :—_E_—3; : g, so an equation 13 y + 2 = 3(30 + 3). The function is f(z)=gx—§,—3gx36. . We need to solve the given equation for y. :c + (y — 1)2 = 0 4: (y — 1)2 = -$ 4:) y — 1 : :l: —a: {i} y = 1 :l: t/ —at. The expression with the positive radical represents the top half of the parabola, and the one with the negative radical represents the bottom half. Hence, we want f (w) = 1 — ‘/—:1:. Note that the domain is a; S 0. (w — 1)2 + 3,12 =1 4:} y = :lzt/1 — (m — 1)2 = :tx/2x — $2. The top halfis given by the function f(:1:) = V253 — 032,0 3 m S 2. For —1 S a: g 2, the graph is the line with slope 1 and y-intercept 1, that is, the line y = :3 + 1. For 2 < :1: S 4, the ra h is the line with slo e —§ and w-interce t4 [which corresponds to the oint 4,0 ], so g P P 2 P P $+1 if—igxg2 —0=—§:v—4 =—§$+6.Sothefunctionis :c = y 2( ) 2 16“ —gx+6 if2<mg4 For a: S O, the graph is the line y = 2. For 0 < a: S 1, the graph is the line with slope —2 and y—intercept 2, that is, the line y = —23: + 2. For a: > 1, the graph is the line with slope 1 and x-intercept 1, that is, the line 2 if m S 0 y: 1(m—1)=$—1.Sothefunctionisf(z)= —2w+2 if 0 <1: 3 1. an — 1 if 1 < a: Let the length and width of the rectangle be L and W. Then the perimeter is 2L + 2W 2 20 and the area is A = LW. Solving the first equation for W in terms of L gives W = 20 ; 2L 2 10 — L. Thus, A(L) = L(10 — L) = 10L ~ L2. Since lengths are positive, the domain ofA is 0 < L < 10. If we further restrict L to be larger than W, then 5 < L < 10 would be the domain. Let the length and width of the rectangle be L and W. Then the area is LW 2 16, so that W = 16/ L. The perimeter is P = 2L + 2W, so P(L) = 2L + 2(16/L) = 2L + 32/L, and the domain of P is L > 0, since lengths must be positive quantities. If we further restrict L to be larger than W. then L > 4 would be the domain. ...
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