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Solution of exp equation

  1. Sep 25, 2013 #1
    No. of Real solution of ##2^x-x^2 = 1##

    Solution (Given as):: Clearly ##x = 0## and ##x = 1## are solution of Given equation.

    Formal; define ##f(x) = 2^x - x^2 - 1##, so ##f'(x) = 2^x\ln 2 - 2x##, ##f''(x) = 2^x\ln^2 2 - 2##,

    ##f'''(x) = 2^x\ln^3 2 > 0##. Study the variation, in order to estimate the values of the roots and

    critical points

    Now I Did not Under How can we check Nature of ##f(x)## using ##f^{'''}(x)##

    Please Help me

    Thanked
     
  2. jcsd
  3. Sep 26, 2013 #2

    UltrafastPED

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    Note that f'''(x) is always positive. This tells you something about f''(x).

    Draw a simple graph of these derivatives.
     
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