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Math Help - system of equations

  1. #1
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    system of equations

    First off my last question was answered in such an awesome way! I am so thankful for this site- You see I'm in the hospital and am trying to complete my math homework for my class via laptop, I only have minimal calculator functions on this thing and no book to reference the how to's- so I can understand it. The hw is do monday regardless of my condition as says my teacher so this site is a heaven sent. Anyway need help with this one:

    y= x^3-x
    y= e^x

    solve the system of equations using any method
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  2. #2
    Member Jonboy's Avatar
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    i just want to make sure the exponents are correct here.

    the first equation is:

    y = x^{3} - x

    right?
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  3. #3
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    yes the first has an exponent of 3 and the second is e to the x power
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by racersteve47 View Post
    First off my last question was answered in such an awesome way! I am so thankful for this site- You see I'm in the hospital and am trying to complete my math homework for my class via laptop, I only have minimal calculator functions on this thing and no book to reference the how to's- so I can understand it. The hw is do monday regardless of my condition as says my teacher so this site is a heaven sent. Anyway need help with this one:

    y= x^3-x
    y= e^x

    solve the system of equations using any method
    First seeing that there are two solutions to this equation, we first note that

    if f(x)=x^3-x-e^x

    f(2)<0 and f(2.5)>0

    Thus since this function is continuous we can see by the Intermediate Value Theorem that

    \exists{c}\in(2,2.5)\backepsilon~f(c)=0

    Using the Intermediate Value Theorem again but this time for the second zero we see that

    \exists{c}\in(4,4.5)\backepsilon~f(c)=0

    So now using 2.25 and 4.25 as intital guesses in Newton's method we see that

    \text{As }n\to\infty~~f_n(x_n)\to2.3217

    and

    f_{n_1}(x_{n_1})\to{4.3717} respectively


    Therefore f(4.3717)\approx{f(2.3217)}\approx{0}
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