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

[shrody]

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?

 

[HallsofIvy]

German? You are putting all your verbs at the end of the sentence!

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

 

[Architeuthis Dux]

1. The preimage of a function is the set of all inputs that produce a given output. In simpler terms, it is the set of values that, when plugged into the function, will result in the desired output. For example, if we have a function f(x) = x^2, the preimage of 4 would be {-2, 2} because both -2 and 2, when squared, produce an output of 4. So, the preimage of a function is essentially the inverse of the function, where the input and output are swapped.

2. To solve for the preimage of the function x-1=e^-x, we need to find the set of values for x that will produce a given output of e^-x. In this case, the output is e^-x and the given interval is [-2,2]. To find the preimage, we can simply plug in values from the interval and see which ones produce the desired output. For example, when x=-1, the equation becomes -1-1=e^-(-1) which simplifies to -2=e, which is not true. But when x=0, the equation becomes 0-1=e^-0 which simplifies to -1=1, which is also not true. So, we can see that there is no solution in this interval that will produce the desired output.

In conclusion, the function x-1=e^-x does not have a solution in the interval [-2,2]. We can verify this by graphing the function and seeing that it does not intersect with the y=e^-x curve in this interval. I hope this helps to clarify the concept of preimage and how to solve for it in a given interval.