For the function y=(e^2x)+(3x)-(10), use Newton's method and the calculator method to find the x value for which y=15. Please show your work so that I can understand the question! Thank you so much!! it means alot!!

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  1. so you are solving
    e^(2x) + 3x - 25 = 0

    let y = e^(2x) + 3x - 25
    y' = 2e^(2x) + 3

    newtons's formula says
    X = x - f(x)/f'(x)

    where x is your starting x value and X is the new value.
    Hopefully your X will approach x, when that happens you have the solution.

    X = x - (e^(2x) + 3x - 25)/(2e^(2x) + 3)
    = (2xe^(2x) - e^(2x) + 25)/(2e^2(2x) + 3)

    I made a rough sketch of y = e^(2x) + 3x - 25
    and noticed that there was an x-intercept, which would be your solution, at appr x = 1.5
    So I will make that my starting x
    X ---- x
    -------- 1.5
    1.5096 ---1.50915695
    1.50951695 -- 1.509515688
    1.509515688 -- 1.50951688

    Wow, got the answer correct to 9 decimal places after only 3 iterations.

    The key thing is that you start with an initial guess close to the real number.

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