Exercises 1.2–1010–0.50.5C=2C=1C=0.5C=0C=−0.5C=−1C=−2Figure 1–A: Graphs of the functionsy=Ce3x+1forC=0,±0.5,±1, and±2.17.DiFerentiatingφ(x), we ±nd thatφ0(x)=±21−cex²0=³2(1−cex)−1´0=2(−1) (1−cex)−2(1−cex)0cex(1−cex)−2.(1.2)On the other hand, substitution ofφ(x)foryinto the right-hand side of the given equationyieldsφ(x)(φ(x)−2)2=1221−cex±21−cex−2²=21−cex±11−cex−1²=21−cex1−(1−cex)1−cex=2cex(1−cex)2,which is identical toφ0(x) found in (1.2).
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.