show that the function f(x) = x - e^-x must have atleast one root 0<=x<=1

maybe im just dumb but the whole *at least one* part has me a bit confused

would i just show that f'(x) is positive and increasing? and plug in 0 and 1 into f'(x)?

yes, just test the end points.
If the endpoints have a different sign and the function is continuous, then you know the function crosses the x-axis at least once

E. hint: you know the function is continuous because you are adding (or subtracting) 2 continuous functions (x and e^-x)

This post was edited by TritonV8 on Apr 8 2013 10:01pm

yay for instinct!

f(x) = g(x) - h(x)
g(x) = x
h(x) = e^(-x)

g(x) and h(x) are clearly continuous, therefore g(x) - h(x) is clearly continuous, therefore f(x) is continuous (if you want to go more formal use Weierstrass's definition)

then using the following 2 results with the intermediate value theorem, there must be at least one value for x (between 0 and 1) that will return 0 for the function.
f(0) = -1
f(1) ≈ 0.6321

/e
beaten by TritonV8

This post was edited by Azrad on Apr 8 2013 10:19pm

still helpful