Derivative of Piece-Wise Function

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SUMMARY

The discussion focuses on the differentiability of piece-wise functions, specifically analyzing the function f(x) defined as x^3 for x ≤ 1 and 3x for x > 1. It concludes that f(x) is not differentiable at x=1 due to a discontinuity, despite both one-sided derivatives being equal to 3. Additionally, the second piece-wise function f(x) defined as x^2 for x ≤ 1 and 2x for x > 1 is shown to be differentiable at x=1, as both left-hand and right-hand derivatives exist and are equal to 2.

PREREQUISITES
  • Understanding of piece-wise functions
  • Knowledge of one-sided derivatives
  • Familiarity with limits and continuity
  • Basic calculus concepts, including differentiation
NEXT STEPS
  • Study the concept of continuity in piece-wise functions
  • Learn about the formal definition of derivatives
  • Explore examples of non-differentiable points in calculus
  • Investigate the implications of differentiability on graph behavior
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and piece-wise functions, as well as educators seeking to clarify concepts of continuity and differentiability.

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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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