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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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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. Evaluate at 0 and 1. What can you conclude? –  Rankeya Oct 19 '12 at 20:11
    
if we use derivative test we get that,$e^x+1>0$,so only solution is complex space right?because $e^x$ never equal to $-1$ –  dato datuashvili Oct 19 '12 at 20:14
    
No, the derivative test shows that the function is monotonically increasing. You are not trying to find the extremum points in this question. –  Rankeya Oct 19 '12 at 20:16
    
yes you are right and it is what i have said,for find extremum,we must set equal $e^x+1=0$ ,it is what i have said in my previous comment,please look at it –  dato datuashvili Oct 19 '12 at 20:17
    
What I don't understand is why you are trying to find the extremum points of the function to locate its zeroes. –  Rankeya Oct 19 '12 at 20:19

2 Answers 2

up vote 2 down vote accepted

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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they never intersect it on real number space –  dato datuashvili Oct 19 '12 at 20:15
    
@dato please see the wolframalpha link above. It has only one solution. –  vesszabo Oct 19 '12 at 20:25
    
@dato : I think you should plot them instead of saying they never intersect, because clearly you haven't. LJym89's suggestion is very clever and allows you to see immediately that there will be a unique zero. Using the mean value theorem you can show immediately that this zero will be between $0$ and $1$. And then you will be happy. The solution is real. =) –  Patrick Da Silva Oct 19 '12 at 20:26
    
yes and it is not real solution right –  dato datuashvili Oct 19 '12 at 20:26
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Dear @dato: You figured out the function is montonically increasing in $\mathbb{R}$. Then it must be injective in $\mathbb{R}$. This means that is can take a value at most once. You figured out there is one zero in the interval $[0,1]$. Combining these observations you get that this must be the only zero. –  Rankeya Oct 19 '12 at 20:40

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