16_Cal_Solution of Calculus_6e

16_Cal_Solution of - -2-1121234C01S02.048:Given:f(x =|x−3| Ifx=0 thenf(x =x−3 so the graph offconsists of the straightline through(3,0 with

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Unformatted text preview: -2-1121234C01S02.048:Given:f(x) =|x−3|. Ifx=0 thenf(x) =x−3, so the graph offconsists of the straightline through (3,0) with slope 1 forx=3. Ifx <0 thenf(x) =−x+ 3, so the graph offconsists of thatpart of the straight line with slope−3 andy-intercept (0,3). These two line segments fit together perfectlyat the point (3,0); there is no break or gap or discontinuity in the graph off.C01S02.049:Given:f(x) =|2x+ 5|. The two cases are determined by the point where 2x+ 5 changessign, which is wherex=−52. Ifx=−52, thenf(x) = 2x+ 5, so the graph offconsists of the part of theline with slope 2 andy-intercept 5 for whichx=−52. Ifx <−52, then the graph offis the part of thestraight liney=−2x−5 for whichx <−52.C01S02.050:The graph consists of the part of the liney=−xfor whichx <0 together with the part ofthe parabolay=x2for whichx=0. The two graphs fit together perfectly at the point (0,0); there is nobreak, gap, jump, or discontinuity there. The graph is shown next....
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This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.

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