For the following exercises (1-5), write Newton\u2019s formula as [latex]x_{n+1}=F(x_n)[\/latex] for solving [latex]f(x)=0[\/latex].<\/p>\r\n\r\n
1.<\/strong> [latex]f(x)=x^2+1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 2.<\/strong> [latex]f(x)=x^3+2x+1[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165042562970\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042562970\"]\r\n [latex]F(x_n)=x_n-\\frac{(x_n)^3+2x_n+1}{3(x_n)^2+2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 3.<\/strong> [latex]f(x)= \\sin x[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 4.<\/strong> [latex]f(x)=e^x[\/latex]<\/p>\r\n[reveal-answer q=\"210817\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"210817\"][latex]F(x_n)=x_n-\\frac{e^{x_n}}{e^{x_n}}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 5.<\/strong> [latex]f(x)=x^3+3xe^x[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n For the following exercises (6-8), solve [latex]f(x)=0[\/latex] using the iteration [latex]x_{n+1}=x_n-cf(x_n)[\/latex], which differs slightly from Newton\u2019s method. Find a [latex]c[\/latex] that works and a [latex]c[\/latex] that fails to converge, with the exception of [latex]c=0[\/latex].<\/p>\r\n\r\n 6.<\/strong> [latex]f(x)=x^2-4[\/latex], with [latex]x_0=0[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165042910261\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042910261\"]\r\n [latex]|c|>0.5[\/latex] fails, [latex]|c|\\le 0.5[\/latex] works<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 7.<\/strong> [latex]f(x)=x^2-4x+3[\/latex], with [latex]x_0=2[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 8.<\/strong> What is the value of \"[latex]c[\/latex]\" for Newton\u2019s method?<\/p>\r\n[reveal-answer q=\"fs-id1165043249487\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043249487\"]\r\n [latex]c=\\frac{1}{f^{\\prime}(x_n)}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n For the following exercises (9-16), start at<\/p>\r\n a. [latex]x_0=0.6[\/latex] and<\/p>\r\n b. [latex]x_0=2[\/latex].<\/p>\r\n Compute [latex]x_1[\/latex] and [latex]x_2[\/latex] using the specified iterative method.<\/p>\r\n\r\n 9.<\/strong> [latex]x_{n+1}=(x_n)^2-\\frac{1}{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 10.<\/strong> [latex]x_{n+1}=2x_n(1-x_n)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165042376400\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042376400\"]\r\n a. [latex]x_1=\\frac{12}{25}, \\, x_2=\\frac{312}{625}[\/latex]; b. [latex]x_1=-4, \\, x_2=-40[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 11.\u00a0<\/strong>[latex]x_{n+1}=\\sqrt{x_n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 12.\u00a0<\/strong>[latex]x_{n+1}=\\frac{1}{\\sqrt{x_n}}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165042318572\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042318572\"]\r\n a. [latex]x_1=1.291, \\, x_2=0.8801[\/latex]; b. [latex]x_1=0.7071, \\, x_2=1.189[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 13.\u00a0<\/strong>[latex]x_{n+1}=3x_n(1-x_n)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 14.\u00a0<\/strong>[latex]x_{n+1}=(x_n)^2+x_n-2[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165043427460\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043427460\"]\r\n a. [latex]x_1=-\\frac{26}{25}, \\, x_2=-\\frac{1224}{625}[\/latex]; b. [latex]x_1=4, \\, x_2=18[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 15.\u00a0<\/strong>[latex]x_{n+1}=\\frac{1}{2}x_n-1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 16.\u00a0<\/strong>[latex]x_{n+1}=|x_n|[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165043253496\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043253496\"]\r\n a. [latex]x_1=\\frac{6}{10}, \\, x_2=\\frac{6}{10}[\/latex]; b. [latex]x_1=2, \\, x_2=2[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n For the following exercises (17-26), solve to four decimal places using Newton\u2019s method and a computer or calculator. Choose any initial guess [latex]x_0[\/latex] that is not the exact root.<\/p>\r\n\r\n 17.\u00a0<\/strong>[latex]x^2-10=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 18.\u00a0<\/strong>[latex]x^4-100=0[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165042333506\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042333506\"]\r\n 3.1623 or -3.1623<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 19.\u00a0<\/strong>[latex]x^2-x=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 20.<\/strong>[latex]x^3-x=0[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165043257524\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043257524\"]\r\n 0, -1, or 1<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 21.\u00a0<\/strong>[latex]x+5 \\cos (x)=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 0<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 23.\u00a0<\/strong>[latex]\\frac{1}{1-x}=2[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 24.\u00a0<\/strong>[latex]1+x+x^2+x^3+x^4=2[\/latex]<\/p>\r\n[reveal-answer q=\"976827\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"976827\"]0.5188 or -1.2906[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 25.\u00a0<\/strong>[latex]x^3+(x+1)^3=10^3[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 26.\u00a0<\/strong>[latex]x= \\sin^2 (x)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165042604695\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042604695\"]\r\n 0<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n For the following exercises (27-30), use Newton\u2019s method to find the fixed points of the function where [latex]f(x)=x[\/latex]; round to three decimals.<\/p>\r\n\r\n 27.\u00a0<\/strong>[latex] \\sin x[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 28.\u00a0<\/strong>[latex] \\tan (x)[\/latex] on [latex]x \\in (\\frac{\\pi }{2},\\frac{3\\pi }{2})[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165043348635\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043348635\"]4.493[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 29.\u00a0<\/strong>[latex]e^x-2[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 30.\u00a0<\/strong>[latex]\\ln (x)+2[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165043395328\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043395328\"]\r\n 0.159, 3.146<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n Newton\u2019s method can be used to find maxima and minima of functions in addition to the roots. In this case apply Newton\u2019s method to the derivative function [latex]f^{\\prime}(x)[\/latex] to find its roots, instead of the original function. For the following exercises (31-32), consider the formulation of the method.<\/p>\r\n\r\n 31.\u00a0<\/strong>To find candidates for maxima and minima, we need to find the critical points [latex]f^{\\prime}(x)=0[\/latex]. Show that to solve for the critical points of a function [latex]f(x)[\/latex], Newton\u2019s method is given by [latex]x_{n+1}=x_n-\\frac{f^{\\prime}(x_n)}{f^{\\prime \\prime}(x_n)}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 32.\u00a0<\/strong>What additional restrictions are necessary on the function [latex]f[\/latex]?<\/p>\r\n[reveal-answer q=\"fs-id1165042350213\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042350213\"]\r\n We need [latex]f[\/latex] to be twice continuously differentiable.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n For the following exercises (33-40), use Newton\u2019s method to find the location of the local minima and\/or maxima of the following functions; round to three decimals.<\/p>\r\n\r\n 33.\u00a0<\/strong>Minimum of [latex]f(x)=x^2+2x+4[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 34.\u00a0<\/strong>Minimum of [latex]f(x)=3x^3+2x^2-16[\/latex]<\/p>\r\n[reveal-answer q=\"949157\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"949157\"][latex]x=0[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 36.\u00a0<\/strong>Maximum of [latex]f(x)=x+\\frac{1}{x}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165043248810\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043248810\"]\r\n [latex]x=-1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 37.\u00a0<\/strong>Maximum of [latex]f(x)=x^3+10x^2+15x-2[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 38.\u00a0<\/strong>Maximum of [latex]f(x)=\\frac{\\sqrt{x}-\\sqrt[3]{x}}{x}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165043431659\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043431659\"]\r\n [latex]x=5.619[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 40.\u00a0<\/strong>Minimum of [latex]f(x)=x^4+x^3+3x^2+12x+6[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165043427422\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043427422\"]\r\n [latex]x=-1.326[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n For the following exercises (41-44), use the specified method to solve the equation. If it does not work, explain why it does not work.<\/p>\r\n\r\n 41.\u00a0<\/strong>Newton\u2019s method, [latex]x^2+2=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 42.\u00a0<\/strong>Newton\u2019s method, [latex]0=e^x[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165042374581\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042374581\"]\r\n There is no solution to the equation.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 43.\u00a0<\/strong>Newton\u2019s method, [latex]0=1+x^2[\/latex] starting at [latex]x_0=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n It enters a cycle.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n For the following exercises (45-49), use the secant method<\/span>, an alternative iterative method to Newton\u2019s method. The formula is given by<\/p>\r\n\r\n 45.\u00a0<\/strong>Find a root to [latex]0=x^2-x-3[\/latex] accurate to three decimal places.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 46.\u00a0<\/strong>Find a root to [latex]0= \\sin x+3x[\/latex] accurate to four decimal places.<\/p>\r\n[reveal-answer q=\"fs-id1165043094069\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043094069\"]\r\n 0<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 47.\u00a0<\/strong>Find a root to [latex]0=e^x-2[\/latex] accurate to four decimal places.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 48.\u00a0<\/strong>Find a root to [latex]\\ln (x+2)=\\frac{1}{2}[\/latex] accurate to four decimal places.<\/p>\r\n[reveal-answer q=\"fs-id1165042479712\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042479712\"]\r\n -0.3513<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 49.\u00a0<\/strong>Why would you use the secant method over Newton\u2019s method? What are the necessary restrictions on [latex]f[\/latex]?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n For the following exercises (50-54), use both Newton\u2019s method and the secant method to calculate a root for the following equations. Use a calculator or computer to calculate how many iterations of each are needed to reach within three decimal places of the exact answer. For the secant method, use the first guess from Newton\u2019s method.<\/p>\r\n\r\n 50.\u00a0<\/strong>[latex]f(x)=x^2+2x+1, \\, x_0=1[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165042473414\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042473414\"]\r\n Newton: 11 iterations, secant: 16 iterations<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 51.\u00a0<\/strong>[latex]f(x)=x^2, \\, x_0=1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 52. <\/strong>[latex]f(x)= \\sin x, \\, x_0=1[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165042364336\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042364336\"]\r\n Newton: three iterations, secant: six iterations<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 54.\u00a0<\/strong>[latex]f(x)=x^3+2x+4, \\, x_0=0[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165042681162\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042681162\"]\r\n Newton: five iterations, secant: eight iterations<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n In the following exercises (55-56), consider Kepler\u2019s equation regarding planetary orbits, [latex]M=E-\\varepsilon \\sin (E)[\/latex], where [latex]M[\/latex] is the mean anomaly, [latex]E[\/latex] is eccentric anomaly, and [latex]\\varepsilon [\/latex] measures eccentricity.<\/p>\r\n\r\n 55.\u00a0<\/strong>Use Newton\u2019s method to solve for the eccentric anomaly [latex]E[\/latex] when the mean anomaly [latex]M=\\frac{\\pi }{3}[\/latex] and the eccentricity of the orbit [latex]\\varepsilon =0.25[\/latex]; round to three decimals.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 56.\u00a0<\/strong>Use Newton\u2019s method to solve for the eccentric anomaly [latex]E[\/latex] when the mean anomaly [latex]M=\\frac{3\\pi }{2}[\/latex] and the eccentricity of the orbit [latex]\\varepsilon =0.8[\/latex]; round to three decimals.<\/p>\r\n\r\n The following two exercises consider a bank investment (57-58). The initial investment is [latex]$10,000[\/latex]. After 25 years, the investment has tripled to [latex]$30,000[\/latex].<\/p>\r\n\r\n 57.\u00a0<\/strong>Use Newton\u2019s method to determine the interest rate if the interest was compounded annually.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n 58.\u00a0<\/strong>Use Newton\u2019s method to determine the interest rate if the interest was compounded continuously.<\/p>\r\n[reveal-answer q=\"fs-id1165043317194\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043317194\"]\r\n 4.394%<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n 59.\u00a0<\/strong>The cost for printing a book can be given by the equation [latex]C(x)=1000+12x+(\\frac{1}{2})x^{2\/3}[\/latex]. Use Newton\u2019s method to find the break-even point if the printer sells each book for [latex]$20[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>","rendered":" For the following exercises (1-5), write Newton\u2019s formula as [latex]x_{n+1}=F(x_n)[\/latex] for solving [latex]f(x)=0[\/latex].<\/p>\n 1.<\/strong> [latex]f(x)=x^2+1[\/latex]<\/p>\n<\/div>\n<\/div>\n 2.<\/strong> [latex]f(x)=x^3+2x+1[\/latex]<\/p>\n [latex]F(x_n)=x_n-\\frac{(x_n)^3+2x_n+1}{3(x_n)^2+2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n 3.<\/strong> [latex]f(x)= \\sin x[\/latex]<\/p>\n<\/div>\n<\/div>\n 4.<\/strong> [latex]f(x)=e^x[\/latex]<\/p>\n