Absolute value and piecewise functions

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Homework Help Overview

The discussion revolves around determining the continuity and differentiability of a piecewise function defined as f(x) = x² + 3 for |x| ≤ 1 and f(x) = |x| + 3 for |x| > 1. Participants are exploring the differentiability aspect, particularly at the points where the function changes its definition.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to verify continuity by evaluating the function at specific points and considering limits from both sides. There is also a discussion about rewriting the piecewise function correctly and analyzing the behavior at critical points like x = 1 and x = -1.

Discussion Status

Some participants have provided clarifications regarding the correct formulation of the piecewise function and suggested focusing on the critical points separately. There is an ongoing exploration of the limits and differentiability without reaching a consensus on the final interpretation.

Contextual Notes

Participants are questioning the assumptions made in the rewriting of the piecewise function and the implications for differentiability at the transition points. There is also a mention of a final question regarding the transformation of an absolute value expression.

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Homework Statement



Find if continuous and differentiable. I am having problems with the differentiable part.


Homework Equations



f(x) = x² + 3, |x| ≤ 1
f(x) = |x| + 3, |x| > 1

The Attempt at a Solution



(1)^2 + 3 = 4
|1| + 3 = 4
∴ It is continuous

Now, rewriting it, you have...
x² + 3, x ≥ -1 or x ≤ 1
-x + 3, x < -1 or x > 1
x + 3, x < -1 or x > 1

Finding the limits from left and right of 1...
From the left = 2x, = 2
From the right: can be both -1 and 1...

Where did I go wrong?
 
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939 said:

Homework Statement



Find if continuous and differentiable. I am having problems with the differentiable part.


Homework Equations



f(x) = x² + 3, |x| ≤ 1
f(x) = |x| + 3, |x| > 1

The Attempt at a Solution



(1)^2 + 3 = 4
|1| + 3 = 4
∴ It is continuous

Now, rewriting it, you have...
x² + 3, x ≥ -1 or x ≤ 1
-x + 3, x < -1 or x > 1
x + 3, x < -1 or x > 1

Finding the limits from left and right of 1...
From the left = 2x, = 2
From the right: can be both -1 and 1...

Where did I go wrong?

No, you've got
x² + 3, x ≥ -1 or x ≤ 1
-x + 3, x < -1
x + 3, x > 1

Worry about the two point x=1 and x=(-1) separately.
 
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Dick said:
No, you've got
x² + 3, x ≥ -1 or x ≤ 1
-x + 3, x < -1
x + 3, x > 1

Worry about the two point x=1 and x=(-1) separately.

Thanks a lot!

Final question can |(2/3)(x)| be turned into (2/3)(-x) and (2/3)(x)?
 
939 said:
Thanks a lot!

Final question can |(2/3)(x)| be turned into (2/3)(-x) and (2/3)(x)?

Sure. |2x/3| is -2x/3 if x<0 and 2x/3 if x>=0.
 
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Likes   Reactions: 1 person

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