Solution Manual For Single Variable Calculus, 8th Edition
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1FUNCTIONS AND MODELS1.1Four Ways to Represent a Function1.The functions() =+√2−and() =+√2−give exactly the same output values for every input value, soandare equal.2.() =2−−1=(−1)−1=for−16= 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 about−02, 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 is−2≤≤4, or[−24].The range ofconsists of all-values on the graph of. For this function, the range is−1≤≤3, or[−13].(f ) Asincreases from−2to1,increases from−1to3. Thus,is increasing on the interval[−21].4.(a) The point(−4−2)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 ofare−2and2.(c)() =−1is equivalent to=−1. When=−1, we have=−3and= 4.(d) Asincreases from0to4,decreases from3to−1. Thus,is decreasing on the interval[04].(e) The domain ofconsists of all-values on the graph of. For this function, the domain is−4≤≤4, or[−44].The range ofconsists of all-values on the graph of. For this function, the range is−2≤≤3, 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( ) = (12−85). The highest point occurs at about(17115).Thus, the range of the vertical ground acceleration is−85≤≤115. 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{|0≤≤2}, whereis measured in hours. The rangeof the function is{|0≤≤120}, whereis measured in miles.uplicible9
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