Solve Preimage Function & Find x-1=e^-x Solution [-2,2]

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The discussion focuses on understanding the preimage of a function and solving the equation x - 1 = e^(-x) within the interval [-2, 2]. A participant seeks clarification on the concept of a preimage with a simple example. The equation is evaluated at specific points, revealing that at x = 0, x - 1 equals -1 while e^(-x) is greater than -1, and at x = 2, x - 1 equals 1 while e^(-2) is less than 1. The conversation highlights the importance of analyzing function behavior at defined points to determine solutions within the specified interval. Overall, the thread emphasizes the need for clarity in mathematical concepts and problem-solving approaches.
shrody
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1.My first problem revolves around the preimage of a function. I have never truly understood this concept, if anyone can clear this with a simple example I would appreciate it.
2.Does x-1=e^-x a solution in the interval [-2, 2] have?
 
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German? You are putting all your verbs at the end of the sentence!:devil:

It's pretty easy to see, isn't it, that when x= 0 x-1= -1 and e^{-x}= e^{-1}> -1 while if x= 2, x-1= 1 and e^{-2}= .1356< 1.
 

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