why does e^(-x) = x have at least one solution?

Question

so sorry, i posted this question like 30 sec ago, but i made a mistake in the expression, so this is the right one...

thank you so much for your help!

Answer 1

Because e^(-x) - x is a continuous funtion, which is positive at -1 and negative at 1, and so by the mean value theorem there must be some point between -1 and 1 such that e^(-x) - x = 0, and this is a solution to e^x = -x.

Edit: the numerical value of the solution is approximately x≈0.56714 32904 09783 87299 99686 62210 36

Answer 2

I would first suggest that you graph both of these equations which will show you that they must cross--only once. In more technical language, since both of these functions are continuous their difference must also be continuous: e^(-x)-x is continuous. Now evaluate this new function at -10 and +10. Since the former is positive and the latter is negative the intermediate value theorem guarantees that the function must have equaled zero somewhere between x=-10 and x=10.

Answer 3

e^(-x) = x, rearrange to form the function

g(x) = e(-x) -x, what value of x will make g(x) = 0?

g'(x) = - e(-x) -1, this is always negative, so is g(x) is a decreasing function, therefore it will have at most only one zero.

we know g(0) = 1
g(0.5) = 0.106
g(0.6) = -0.051

the solution therefore is a number between 0.5 and 0.6, this function has at least one zero because it is a continous function that is decreasing over an interval that maps from positive to negative values.