{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Title" -1 18 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 1 2 2 2 1 1 1 1 }3 1 0 0 12 12 1 0 1 0 2 2 19 1 }{PSTYLE "R3 Font 0" -1 256 1 {CSTYLE "" -1 -1 "He lvetica" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "R3 Font 2" -1 257 1 {CSTYLE "" -1 -1 "Courier" 1 14 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 46 "Making use of derivatives and its applications" }}{PARA 0 "" 0 "" {TEXT -1 34 "Consider the pol ynomial function " }}{PARA 0 "" 0 "" {TEXT -1 79 " \+ p(x) = -(x-1)*(x-2)*(x-3)*(x-4)*(x-5)+5. " }}{PARA 0 "" 0 "" {TEXT -1 11 "Questions: " }}{PARA 0 "" 0 "" {TEXT -1 75 " Us e the following suggested Maple syntax to solve the following question s:" }}{PARA 0 "" 0 "" {TEXT -1 46 "(1) Find the local extreme values of p(x). " }}{PARA 0 "" 0 "" {TEXT -1 132 "(2) Use the graph of th e derivative function dp(x) of p(x) to determine the interval(s) where p(x) is increasing and decreasing." }}{PARA 0 "" 0 "" {TEXT -1 171 "(3) Use the graph of the second derivative function ddp(x) of p(x) \+ to find all intervals in which p(x) is concave up and all intervals \+ in which p(x) is concave down." }}{PARA 0 "" 0 "" {TEXT -1 146 "(4) \+ Plot p(x),dp(x) and ddp(x) altogether and explain (a) the relationsh ip between p(x) and dp(x), (b) the relationship between p(x) and ddp(x )." }}{PARA 0 "" 0 "" {TEXT -1 10 " " }}{PARA 0 "" 0 "" {TEXT -1 24 "GUIDE TO THE SOLUTION: " }}{PARA 0 "" 0 "" {TEXT -1 54 " (1) Since p(x)= -(x-1)*(x-2)*(x-3)*(x-4)*(x-5)+5 is" }}{PARA 0 "" 0 "" {TEXT -1 60 "a polynomial function, p is differentiable everywhe re. The" }}{PARA 0 "" 0 "" {TEXT -1 66 "following Maple V segment det ermines the extreme values. The first" }}{PARA 0 "" 0 "" {TEXT -1 59 " thing we do is define p(x) and its first two derivatives." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "p :=x-> \+ -(x-1)*(x-2)*(x-3)*(x-4)*(x-5)+5;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "The following \"dp\" and \"ddp\" are to define the first and secon d derivative functions respectively." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "dp := D(p);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "p1:=x- >diff(p(x),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "p1(x);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "ddp := D(dp);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "p2:=x->diff(p1(x),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "p2(x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "(1) Use the following command to find the critical point s, where we can use them to determine the max or min." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "critpoints := [fsolve(dp(x)=0,x)];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "(2) One way to determine the in tervals in which p(x) is" }}{PARA 0 "" 0 "" {TEXT -1 66 "increasing \+ or decreasing is to plot the graph of p'(x). We start" }}{PARA 0 "" 0 "" {TEXT -1 64 "with a plot of p'(x) over an interval that include s all of the" }}{PARA 0 "" 0 "" {TEXT -1 16 "critical points." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plo t(dp(x),x=0..6,y=-20..30);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 244 "Hi nt: You need to examine the zeroes of dp(x), and if the graph of dp(x) is changing from positive to negative at a point x=a, then we have a \+ relative maximum for p(x) at x=a. Use the similar strategy to determin e the relative minimum for p(x)." }}{PARA 0 "" 0 "" {TEXT -1 57 "(3) \+ Since p'(x) is a polynomial the critical points of" }}{PARA 0 "" 0 " " {TEXT -1 69 " p'(x) are the zeros of the second derivative p''(x). Thus we find" }}{PARA 0 "" 0 "" {TEXT -1 35 "the zeros of the second derivative." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "solve(ddp(x)=0,x);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "plot(ddp(x),x=0..6,y=-20.. 30);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 258 "Notice that if ddp(x) ch anges sign at x=a from positive to negative, then the function p(x) c hanges from concave upward to concave downward at x=a. Use the similar strategy to determine the interval(s) where p(x) is changing from con cave downward to upward." }}{PARA 0 "" 0 "" {TEXT -1 111 "(4) Use the \+ following command to plot p(x), dp(x) and ddp(x) together. You need to indicate which one is which." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "plot(\{p(x),dp(x),ddp(x)\},x=0. .6,y=-30..30,thickness=3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 29 "Example 1. page 309, number 2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 24 "f:=x->2*Pi*x^2+(2000/x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "simplify(f(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "f1:=x->diff(f(x),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "f1(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "s implify(f1(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "fsolve(f 1(x)=0,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(f(x),x=- 5..5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "plot(\{f(x),f1(x) \},x=-6..6,y=-100..100);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Examp le 2. number 18 on the homework, take a=4 and b=3." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "A:=x->2*x*2*sqrt((144-9*x^2)/16);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "A1:=x->diff(A(x),x);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "A1(x);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 18 "simplify(A1(x),x);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 18 "fsolve(A1(x)=0,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(A1(x),x=0..5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "plot(A1(x),x=0..5,y=-100..100);" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 43 "We conclude that A has a maximum when x = " } {XPPEDIT 18 0 "2.828427125;" "6#$\"+DrUGG!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "plot(\{A(x),A1(x)\},x=0..5,y=-100..100);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "Example 3. homework section 4.6, n umber 20. set height =4 units and radius = 5 units." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "V:=x->Pi*x^2*(4-x*4/5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "V1:=x->diff(V(x),x);" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 16 "fsolve(V1(x),x);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 19 "plot(V1(x),x=0..5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "plot(\{V(x),V1(x)\},x=0..5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 46 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }