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chapter 1 5

chapter 1 5 - 22 23 24 25 26 27 28 29 30 31 SECTION 1.1...

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Unformatted text preview: 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION D x 2+h 2+h m+h _ =_=_ h =——, d f(x)_x+1’s°f(2+h) 2+h+1 3+h’ (33+) ac+h+1an f<w+h)—f(x)_J—zihirﬁi_(w+h)<w+1>—x<x+h+1)= 1 . h _ h _ h(x+h+1)(w+1) (a:+h+1)(\$+1) f(ac) = :c/(Bm — 1) is deﬁned for all :17 except when 0 : 3m — 1 (1} m = , so the domain is {we me e} = (—ooe) u (em)- f(m) = (5x + 4) /(a:2 + 3x + 2) is deﬁned for all m except when 0 = :52 + 3:): + 2 <:> 0 = (x + 2)(:c +1) (e m = —2 or ~1, so the domain is {x E R |w 75 ~2, —1} = (—00, —2) U (—2, —1) U (—1,oo). ooh-I f (t) = x/E + \S/E is deﬁned when t 2 0. These values of t give real number results for x/i, whereas any value oft gives a real number result for 5/7? The domain is [0, oo). g(u) = ﬁ+ V4 — uis deﬁned when u 2 O and 4 — u 2 0 (i) u S 4. Thus, the domain is Ogug4=[0,4]. h(\$) 2 1 /{‘/:c2 — 5a: is deﬁned when 362 ~ 5m > 0 (it w(m — 5) > 0. Note that 2:2 — 5a: aé 0 since that would result in division by zero. The expression 20(55 ~ 5) is positive if :E < 0 or a: > 5. (See Appendix A for methods for solving inequalities.) Thus, the domain is (—00, O) U (5, 00). h(m):x/4T——av2.Nowy=VAL—ﬁzz2 => y2=4—\$2 4:; x2+y2=4,so the graph is the top half of a circle of radius 2 with center at the origin. The domain is {x I 4 — 7:2 Z 0} = {1: | 4 2 m2} = {m | 2 2 lml} = [—2,2]. From the graph, the range is 0 S y S 2, or [0, 2]. f(x) 2 5 is deﬁned for all real numbers, so the domain is JR, or (—00, oo). YT The graph of f is a horizontal line with y-intercept 5. 5 y : 5 —_q.__ ‘J__, O x F(a:) 2 ﬁx + 3) is deﬁned for all real numbers, so the domain is R, or y‘ (—00, 00). The graph of F is a line with cc—intercept —3 and y—intercept g. f (t) = t2 — (it is deﬁned for all real numbers, so the domain is R, or (—00, 00). The graph of f is a parabola opening upward since the coefﬁcient of t2 is positive. To ﬁnd the t—intercepts, let 3; = 0 and solvefort. 0:152 —6t=t(t—6) ¢ t:0andt:6. The t-coordinate of the vertex is halfway between the t—intercepts, that is, at t = 3. Since f(3) = 32 — 6 . 3 : ~9, the vertex is (3, —9). 5 ...
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