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Math Help - Have I shown that one real number solution exists?

  1. #1
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    Have I shown that one real number solution exists?

    Show that there is exactly one real number x for which  e^x + x = 5. Find the integer part of this number. State any theorems you use in your solution.

    My attempt:

    e^x+x-5 = 0

    f(0) = 1-5 = -4

    f(2) \approx 7+2-5 = 4

    By the intermediate value theorem, the continuous function crosses the x axis. The function is always increasing, therefore it crosses the x axis only once.

    Is there an algebraic way for finding the integer part? I just subbed 1 into the original equation and got an answer < 5 then subbed in 2 and got an answer > 5 and deduced that x must be 1.something \therefore the integer part is 1.
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  2. #2
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    Re: Have I shown that one real number solution exists?

    Quote Originally Posted by terrorsquid View Post
    Show that there is exactly one real number x for which  e^x + x = 5. Find the integer part of this number. State any theorems you use in your solution.

    My attempt:

    e^x+x-5 = 0

    f(0) = 1-5 = -4

    f(2) \approx 7+2-5 = 4

    By the intermediate value theorem, the continuous function crosses the x axis. The function is always increasing, therefore it crosses the x axis only once.

    Is there an algebraic way for finding the integer part? I just subbed 1 into the original equation and got an answer < 5 then subbed in 2 and got an answer > 5 and deduced that x must be 1.something \therefore the integer part is 1.
    You haven't shown yet that there's EXACTLY one solution though...
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    Re: Have I shown that one real number solution exists?

    Where did you demonstrate "only increasing"?

    Define "algebraic". Iterative techniques are required.
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    Re: Have I shown that one real number solution exists?

    f'(x) = e^x + 1 which is positive for any value of x, therefore it is always increasing. Is that sufficient?

    Originally I just noticed that the function was basically a shifted version of the e^x function and deemed it as always increasing Not very thorough.
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    Re: Have I shown that one real number solution exists?

    he's left out some steps, but if you accept that the function is always increasing, then there must be exactly one zero.

    @terrorsquid: it would be prudent to show that f'(x) > 0, for all x (EDIT: i didn't see your latest post).
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