Solution Manual For Single Variable Calculus, 8th Edition

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1FUNCTIONS AND MODELS1.1Four Ways to Represent a Function1.The functions() =+2and() =+2give exactly the same output values for every input value, soandare equal.2.() =21=(1)1=for16= 0, soand[where() =] are not equal because(1)is undefined and(1) = 1.3.(a) The point(13)is on the graph of, so(1) = 3.(b) When=1,is about02, so(1)≈ −02.(c)() = 1is equivalent to= 1When= 1, we have= 0and= 3.(d) A reasonable estimate forwhen= 0is=08.(e) The domain ofconsists of all-values on the graph of. For this function, the domain is24, or[24].The range ofconsists of all-values on the graph of. For this function, the range is13, or[13].(f ) Asincreases from2to1,increases from1to3. Thus,is increasing on the interval[21].4.(a) The point(42)is on the graph of, so(4) =2. The point(34)is on the graph of, so(3) = 4.(b) We are looking for the values offor which the-values are equal. The-values forandare equal at the points(21)and(22), so the desired values ofare2and2.(c)() =1is equivalent to=1. When=1, we have=3and= 4.(d) Asincreases from0to4,decreases from3to1. Thus,is decreasing on the interval[04].(e) The domain ofconsists of all-values on the graph of. For this function, the domain is44, or[44].The range ofconsists of all-values on the graph of. For this function, the range is23, or[23].(f ) The domain ofis[43]and the range is[054].5.From Figure 1 in the text, the lowest point occurs at about( ) = (1285). The highest point occurs at about(17115).Thus, the range of the vertical ground acceleration is85115. Written in interval notation, we get[85115].6.Example 1:A car is driven at60mih for2hours. The distancetraveled by the car is a function of the time. The domain of thefunction is{|02}, whereis measured in hours. The rangeof the function is{|0120}, whereis measured in miles.uplicible9

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