Derivative of Piece-Wise Function

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Homework Help Overview

The discussion revolves around the differentiation of piece-wise functions in calculus, specifically focusing on the function defined as f(x) = x^3 for x ≤ 1 and f(x) = 3x for x > 1. Participants are exploring the conditions under which a derivative exists at a point of discontinuity, particularly at x = 1.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are examining the concept of one-sided derivatives and continuity, questioning how the presence of the same slopes from both sides can coexist with a lack of differentiability at a point. They are also discussing the implications of having a "jagged edge" at x = 1 and how that affects the existence of a derivative.

Discussion Status

Some participants suggest that the lack of continuity at x = 1 is a sufficient reason for the function not to be differentiable there. Others are debating whether the existence of equal one-sided derivatives implies differentiability, while also acknowledging the potential for a "hole" in the derivative at that point.

Contextual Notes

There is a focus on the definitions and rules surrounding derivatives, particularly regarding piece-wise functions and the conditions that affect their differentiability. Participants are also considering the implications of continuity and the graphical representation of the functions involved.

Jeremy
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I am doing my calculus homework and two problems are holding me up.

The first says:

Using one-sided derivatives, show that the function f(x) =

x^3, x_<_1
3x, x>1

does not have a derivative at x=1

Now it is painfully obvious that the function is not continuous at x=1. however, i am not entirely sure that is the answer that the book wants. the slopes from both sides seem to be 3, so how can it not have a derivative (aside from the continuity issue)? A similar (but with continuity) problem follows:



Secondly (i will try to answer the ones i have gotten so far):
let f(x) =

x^2, x_<_1
2x, x>1

a) find f'(x) for x<1......i think this is 2x
b) find f'(x) for x>1......2
c) find lim (x-->1-) f'(x)....2
d) find lim (x-->1+) f'(x)....2
e) does lim (x-->1) f'(x) exist? explain
f) use the def to find the left-hand derivative of f at x=1 if it exists...same as (c)=2
g) use the def...right-hand deriv...same as (d)=2
h) does f'(1) exist? explain

according to the rules of derivatives, if the left and right-hand derivatives are the same at a point, then that point has a derivative (assuming continuous). however, it seems to me that, at x=1, there would be a bit of a "jagged edge," somewhat like an absolute value point. therefore, how could a derivative be found?

thanks in advance. feel free to tell me i am horribly wrong in all aspects of my answer.
 
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For the first question, an answer of it isn't differentiable because its not continuous seems to be the answer that would be appropriate to me.

2nd question, yes, it should be differentiable at f'(1). Because the slopes (derivatives) are the same at point x = 1, then it shouldn't have a "jagged egde". If it were graphed it should appear smooth, since the one sided limits of both the functions and derivatives have the same value at x=1.
 
sounds good

thanks a bunch
 
If there is no derivative at x = 1, which there isn't, then there can be no further derivatives at x = 1. There will be a "hole" in the graph of the derivative there. Therefore, the derivative is not differentiable at x = 1, even if both the left-hand and the right-hand derivatives exist and are the same.
 

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