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You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. e^x - 1 = 0 How do I find x?
  • Thread starter Roadrunner2015
  • Start date Apr 14, 2014
R

Roadrunner2015

New member
Joined Apr 14, 2014 Messages 8 I'm finding the critical points of f(x) = ex - x. First step: Find f'(x) f'(x) = ex - 1 Second Step: Set to 0 and solve for x ex = 1 My question now is how to solve for x. x = ln1? If so, where do I go from there? stapel

stapel

Super Moderator
Staff member Joined Feb 4, 2004 Messages 16,582
Roadrunner2015 said: ex = 1 My question now is how to solve for x. x = ln1? If so, where do I go from there? Click to expand...
Where are you needing to "go", now that you have the value of x? ;) pka

pka

Elite Member
Joined Jan 29, 2005 Messages 11,980
Roadrunner2015 said: I'm finding the critical points of f(x) = ex - x. First step: Find f'(x) f'(x) = ex - 1 Second Step: Set to 0 and solve for x ex = 1 My question now is how to solve for x. x = ln1? If so, where do I go from the[e? Click to expand...
\(\displaystyle \ln(1)=~?\) R

Roadrunner2015

New member
Joined Apr 14, 2014 Messages 8
stapel said: Where are you needing to "go", now that you have the value of x? ;) Click to expand...
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? H

HallsofIvy

Elite Member
Joined Jan 27, 2012 Messages 7,763 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"? You must log in or register to reply here. Share: Facebook Twitter Reddit Pinterest Tumblr WhatsApp Email Share Link
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