e^x - 1 = 0 How do I find x?

[Roadrunner2015]

I'm finding the critical points of f(x) = e^x - x.

First step: Find f'(x)

f'(x) = e^x - 1

Second Step: Set to 0 and solve for x

e^x = 1

My question now is how to solve for x.

x = ln1? If so, where do I go from there?

 

[stapel]

e^x = 1

My question now is how to solve for x.

x = ln1? If so, where do I go from there?

Where are you needing to "go", now that you have the value of x?

 

[pka]

I'm finding the critical points of f(x) = e^x - x.
First step: Find f'(x)
f'(x) = e^x - 1
Second Step: Set to 0 and solve for x
e^x = 1
My question now is how to solve for x.
x = ln1? If so, where do I go from the[e?

\(\displaystyle \ln(1)=~?\)

 

[Roadrunner2015]

Where are you needing to "go", now that you have the value of x?

Getting the critical points.

ln(1) = 0 I believe. So plugging in f(0), f(-2), and f(2) I got local min being f(0) and local max being f(2). Would that be correct?

 

[HallsofIvy]

Yes, \(\displaystyle e^0= 1\) (any number to the 0 power is 1) so ln(1)= 0. But where did you get "-2" and "2"?