The equation $$e^{-x}-x+2=0$$ has one root, A.
Find an integer, N, such that
N

Question

The equation $$e^{-x}-x+2=0$$ has one root, A.
Find an integer, N, such that 
N<A<N+1

anonymous · Answer

SOLVING NUMERICALLY

dumbcow · Answer

i would use newtons method

$$\large x_{n+1} =x_{n} -\frac{f(x)}{f'(x)}$$
f'(x) = -e^(-x) -1
initial guess, x_0 = 1

anonymous · Answer

Wait, after the f'(x) = -e^(-x)-1, how do you work with it?

dumbcow · Answer

its a recursive process, start with x=1
then
x = 1- f(1)/f'(1)
then repeat with new x_value until it stops changing hence f(x)=0

anonymous · Answer

But, does e have a value, or what?

dumbcow · Answer

that is to find A.
i guess to find an integer N, you could just plug in values and see when f(x) switches from pos to neg

anonymous · Answer

I'm still confused... Can you show some steps?

dumbcow · Answer

e = 2.72...

dumbcow · Answer

its just a constant

anonymous · Answer

Ok.

dumbcow · Answer

sorry i made it more complicated than it needed to be

plug in 0,1,2,3 into f(x) and see what values you get, then you can guess where the zero is

anonymous · Answer

Hmm Ok. Let me try...

anonymous · Answer

I'm still stuck :(

dumbcow · Answer

f(0) = e^0 -0 +2 = 3

f(1) = e^-1 -1 +2 = 1.37

f(2) = e^-2 -2 +2 = 0.135

f(3) = e^-3 -3+2 = -0.95

dumbcow · Answer

so its decreasing and we can assume crosses x-axis somewhere between 2 and 3

anonymous · Answer

Yeah, the book says 2. Thanks :)