Chapter 1 Solutions cont.

# Chapter 1 Solutions cont. - Section 1.3 Functions and Their...

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Section 1.3 Functions and Their Graphs 19 23. x012 24. y010 y100 25. y 26. y 3 x, x 1 2x, 1 x , x 0 x, 0 x œœ ±Ÿ ² ² Ÿ " x 27. (a) Line through and : y x ab !ß ! "ß " œ Line through and : y x 2 "ß " #ß ! œ ± ³ f(x) x, 0 x 1 x 2, 1 x 2 œ ŸŸ ±³ ²Ÿ œ (b) f(x) 2, x x 2x x œ !Ÿ ²" " Ÿ ² # ß# Ÿ² \$ \$ Ÿ Ÿ % Ú Ý Ý Û Ý Ý Ü 28. (a) Line through 2 and : y x 2 #ß ! œ ± ³ Line through 2 and : m , so y x 2 x a b ß " &ß ! œ œ œ ± œ ± ± ³ " œ ± ³ !±" ±" " " " & &±# \$ \$ \$ \$ \$ f(x) x, 0 x x œ ±³# ²Ÿ # ±³ # & œ "& \$\$ (b) Line through and : m , so y x ±"ß ! !ß ±\$ œ œ ±\$ œ ±\$ ± \$ ±\$±! !±Ð±"Ñ Line through and : m , so y x a b !ß \$ #ß ±" œ œ œ ±# œ ±# ³ \$ ±"±\$ ±% #±! # f(x) x x œ ±\$ ± \$ ±" ² Ÿ ! ±# ³ \$ ! ² Ÿ # œ 29. (a) Line through and : y x a b ±"ß " !ß ! œ ± Line through and : y !ß " "ß " œ " Line through and : m , so y x x a b " ß" \$ ß! œ œ œ± ±" ³"œ± ³ !±" ±" " " " \$ \$±" # # # # # f(x) xx x œ ±± " Ÿ ² ! "! ² Ÿ " " ²² \$ Ú Û Ü "\$ ## (b) Line through and : y x a b ±# ß±" ! œ " # Line through and : y x !ß # "ß ! œ ±# ³ # Line through and : y "ß ±" \$ß ±" œ ±"

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20 Chapter 1 Preliminaries f(x) xx x œ ±# Ÿ Ÿ ! ±# ² # ! ³ Ÿ " ±" " ³ Ÿ \$ Ú Û Ü " # 30. (a) Line through and T : m , so y x 0 x ˆ‰ ˆ ab TT T T T Î # # "±! # # # ß! ß" œ œ œ ± ² œ ±" f(x) , 0 x x, x T œ Ÿ ³ Ÿ ± T T T # # # (b) f(x) A, x Ax T A T x T œ !Ÿ ³ ±ß Ÿ ³ ßŸ ³ Ÿ Ÿ# Ú Ý Ý Ý Û Ý Ý Ý Ü T T T T # # \$ # \$ # 31. (a) From the graph, 1 x ( 2 0) ( ) x4 x # ´² Ê −± ß µ% ß_ (b) 1 1 0 ## ´² Ê ±±´ x 0: 1 0 0 0 ´± ± ´ Ê ´ Ê ´ x 2 x 8 x2 x x (x 4)(x 2) ±± ±² # x 4 since x is positive; Ê´ x 0: 1 0 0 0 ³± ± ´ Ê ³ Ê ³ x 2 x 8 2x 2 x x (x 4)(x 2) # # x 2 since x is negative; Ê³ ± sign of (x 4)(x 2) 2 ïïïïïðïïïïïðïïïïî ²² ± ± % Solution interval: ( 0) ( ) ±#ß µ %ß_ 32. (a) From the graph, x ( 5) ( 1 1) 32 x1 ³Ê ± _ ß ± µ ± ß (b) x 1: 2 Case ³ Ê ´ 3(x 1) ± ² 3x 3 2x 2 x 5. Ê² Ê ³ ± Thus, x ( 5) solves the inequality. − ±_ß± 1x1 : 2 Case ±³ ³ ³ Ê ³ 3(x 1) ± ² 3x 3 2x 2 x 5 which is true ´ ± Ê ´ ± if x 1. Thus, x ( 1 1) solves the − ± ß inequality. 1 x: 3x 3 2x 2 x 5 Case ³³ Ê ² ³ ± Ê ³ ± which is never true if 1 x, so no solution ³ here. In conclusion, x ( 5) ( 1 1).
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Chapter 1 Solutions cont. - Section 1.3 Functions and Their...

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