Why Is the Derivative of e^|x| Undefined at x=0?

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Homework Help Overview

The discussion revolves around differentiating the function \( f(x) = e^{|x|} \) and understanding why the derivative is considered undefined at \( x = 0 \). The subject area includes calculus, specifically the concepts of derivatives and absolute values.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the differentiation of \( e^{|x|} \) by considering the piecewise nature of the absolute value function. They discuss the limits from both sides of \( x = 0 \) and question the behavior of the derivative at that point.

Discussion Status

There is ongoing exploration of the derivative's behavior at \( x = 0 \), with some participants noting the presence of a sharp corner in the graph of \( f(x) \). Guidance has been offered regarding the interpretation of the derivative from both sides of zero, but no consensus has been reached on the implications of this behavior.

Contextual Notes

Participants are grappling with the implications of the piecewise definition of \( |x| \) and how it affects the differentiability of \( f(x) \) at \( x = 0 \). There is a focus on understanding the graphical representation and the concept of a well-defined slope.

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Homework Statement


a)How to differentiate fx=e^|x|?
b)Why when x = 0, f'(x) is undefined

Homework Equations





The Attempt at a Solution


Is it d/dx e^|x| = e^|x|? I have no idea for this question.
 
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|x|=x if x>=0 and |x|=(-x) if x<0. Split the problem into those two cases. To find out what happens AT x=0 look at the limits from both sides.
 


So there are two answer right? -1/e^x and e^x. Why f'x is undefined at x = 0?
 


Hello, please i still don't get it. I sketch 2 graph from the equation and at x = 0, there is a line there. But why f'(x) for e^|x| is undefined when x = 0?
 
takercena said:
Hello, please i still don't get it. I sketch 2 graph from the equation and at x = 0, there is a line there. But why f'(x) for e^|x| is undefined when x = 0?

Hi takercena! :smile:

Your sketch shows two graphs, reflections of each other, joined together …

do they join smoothly, or with a corner? :wink:
 


takercena said:
So there are two answer right? -1/e^x and e^x. Why f'x is undefined at x = 0?

There aren't really two answers, it just that the formula for the answer looks different for x>0 than it does for x<0. If x is close to 0 and negative f'(x)~(-1), if positive then f'(x)~1. There's a sharp corner on the graph of f(x) at x=0, just like on |x|. So f(x) doesn't have a well defined slope there.
 


Oh i see. Thanks
 

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