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suppose we we have following equations and conditions Let $k$ be the number of real solutions of the equation $e^x+x-2=0$ in the interval $[0, 1]$ and and let $n$ be the number of real solutions that are not in $[0,1]$ Which of the following is true?

$k=0$ and $n=1$

$k=1$ and $n=0$

$n=k=1$

$k>1$

$n>1$

first of all what i have tried is this:if we differentiate we get following thing $e^x=-1$ but how it is possible?,using wolfram alpha i got this result

http://www.wolframalpha.com/input/?i=e%5Ex%2Bx-2%3D0

but how to prove it using mathematical procedure?have to i use newtons method for compute actual root or there is some specific theorem which helps me to determine it more easily?

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    generally if we use known theorem that if at bound points function has different sign,then there is at least zero between them,so by we know that in [0...1] there is at least some point where our function is equal to zero,but what about real numbers?2012-10-19

2 Answers 2

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Okay, I am moving my comment as an answer:

Use derivative test to see if the function is increasing or decreasing. A strictly increasing/ decreasing function must be injective, so it can have at most one zero. Note that the function $e^x + x - 2$ takes both positive and negative values by evaluating at 0 and 1. What can you conclude by the intermediate value theorem?

Also, you should not try to find extremum points to solve this question, for the zeroes of $e^x + x -2$ need not be extremum points.

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Consider the functions $f(x) = e^x$ and $g(x) = 2 - x$. Now, the solutions for your equation in the interval $[0,1]$ are the points where $f$ and $g$ intersect. try graphing it!

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    yes Dear @Rankeya,now i now answer of my question2012-10-19