If f+g is differentiable, then are f and g differentiable too?

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Homework Help Overview

The discussion revolves around the differentiability of two continuous functions, f and g, given that their sum, f+g, is differentiable. Participants are exploring whether the differentiability of the sum implies the differentiability of the individual functions.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster asks for a counterexample to demonstrate that f and g may not be differentiable even if their sum is. One participant suggests specific functions, f and g, that are continuous but not differentiable, and discusses their behavior at a particular point.

Discussion Status

Participants are actively engaging with the problem, with one providing a specific example to illustrate their reasoning. There is an ongoing exploration of the implications of continuity and differentiability without reaching a consensus.

Contextual Notes

The original poster requests a counterexample, indicating a focus on understanding the relationship between the differentiability of functions and their sum. The discussion includes considerations of continuity and points where differentiability may fail.

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Homework Statement


if f and g are two continuous functions and f+g is differentiable...are f and g differentiable? if not give a counter example!

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The Attempt at a Solution

 
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I'm thinking not necessarily, but I'm not really sure. Whether a function is differentiable or not depends on whether on not every point on it has a derivative.

So say f is y = x^(1/3), and g is y = -x^(1/3). Both f and g are continuous - the graph of the function is unbroken.

(Technically the definition of continuity is that for every value in the domain the limit from below = limit from above = the value itself but yeah that's not really important.)

f+g is differentiate yeah? f + g just gives y = 0
However, y = x^(1/3) and y = -x^(1/3) are both non differentiable, since at the origin x = 0, the derivative is undefined. You get a divide by zero from memory =P
 


thnx clementc
 


haha no worries! =P
 

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