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Math Help - no. of real solution in exp. equation

  1. #1
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    no. of real solution in exp. equation

    Total no. of real solution of 2^x = 1-x^2


    My Trail:: Let f(x) = 2^x and g(x) = 1-x^2


    Now Using first derivative test, we will check whose slope is Increasing faster rate.


    So f{'}(x) = 2^x\cdot \ln (2) and g^{'}(x) = -2x


    So for The x\leq -1, Slope of f(x) > Slope of g(x)


    So f(x) is always above g(x)


    So no solution for x\leq -1


    and for x \geq  0, Slope of f(x) > Slope of g(x)


    So f(x) is always above g(x)


    So no solution for x > 0


    Now we will calculate in -1 < x < 0


    Now I Did not understand How can I find whether f(x) is g(x) or not


    please help me


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  2. #2
    Forum Admin topsquark's Avatar
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    Re: no. of real solution in exp. equation

    Never underestimate the power of a picture. It will help you refine your solution.

    And you missed an obvious solution: (x, y) = (0, 1).

    -Dan
    Attached Thumbnails Attached Thumbnails no. of real solution in exp. equation-intersection.jpg  
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    Re: no. of real solution in exp. equation

    Thanks moderator


    Yes x = 0 is obivious solution of given equation.


    Now my question is how can I check these two graph intersect in  -1< x < 0 or not.


    and How can we draw figure of these graph in -1<x<0


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    Re: no. of real solution in exp. equation

    Quote Originally Posted by jacks View Post
    Thanks moderator


    Yes x = 0 is obivious solution of given equation.


    Now my question is how can I check these two graph intersect in  -1< x < 0 or not.


    and How can we draw figure of these graph in -1<x<0


    Thanks
    The graph provided by Topskward is the answer to the last line of your post.
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    Re: no. of real solution in exp. equation

    Do you have to show proof (algebraic) ?
    Last edited by chen09; September 26th 2013 at 09:41 PM.
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    Re: no. of real solution in exp. equation

    Quote Originally Posted by chen09 View Post
    Do you have to show proof (algebraic) ?
    It has to be iterative solution
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    Re: no. of real solution in exp. equation

    Quote Originally Posted by votan View Post
    It has to be iterative solution
    I tried Newton-Raphson. It either converges to 1 or it becomes unstable depending on initial guess. I think bisection method should work but I did not put time on it. I found the other root graphically: x = -0.57203.

    2^(-0.57203) = 0.6727
    1 - x^2 = 0.6728
    Last edited by votan; September 27th 2013 at 07:45 AM.
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    Forum Admin topsquark's Avatar
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    Re: no. of real solution in exp. equation

    Quote Originally Posted by votan View Post
    Topskward
    !!!!!!

    -Dan
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    Re: no. of real solution in exp. equation

    Quote Originally Posted by topsquark View Post
    !!!!!!

    -Dan
    I was wandering what is this. I went back to may posts. I am still laughing from it. My apology for mispelling topsquark
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