Help finding the local min in a piecewise function

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Discussion Overview

The discussion revolves around finding the local minimum of a piecewise function defined by two segments: \( f(x) = 16 - x^2 \) for \( -4 \leq x \leq 0 \) and \( f(x) = 2x - 3 \) for \( 0 \leq x \leq 4 \). Participants explore the implications of the first derivative test and the behavior of the function at critical points.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant notes that \( f(0) \) has two values defined, which raises questions about the critical points in the sub-domains.
  • Another participant calculates the derivative of the first piece \( 16 - x^2 \) and finds it equals zero at \( x = 0 \), suggesting it is a "break point".
  • It is mentioned that the derivative of the second piece \( 2x - 3 \) is never zero, indicating no critical points in that segment.
  • Some participants propose that a local minimum must occur at the endpoints \( x = -4 \), \( x = 0 \), or \( x = 4 \), but the ambiguity of \( f(0) \) prevents a definitive conclusion about the local minimum.

Areas of Agreement / Disagreement

Participants express uncertainty about the local minimum due to the conflicting values of \( f(0) \) and the implications of the first derivative test. Multiple competing views remain regarding the identification of critical points and the behavior of the function at those points.

Contextual Notes

The discussion highlights limitations related to the definition of \( f(0) \) and the implications for determining local minima. There is also a lack of consensus on the critical points due to the piecewise nature of the function.

thinkbot
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f(x) = {
{ 16 - x^2, if -4 <= x <= 0
{ 2x - 3, if 0 <= x <= 4
I did the first derivative test and got zero but i don't think that's right. Any help?
 
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Re: Help finding the local min in a peicewise function

One quibble: $$f(0)$$ has two values defined.

Do you find any critical values inside the two sub-domains?
 
Re: Help finding the local min in a peicewise function

f^1(x)={2x
{2
0=2x x=0
thats what i had but i think its wrong
 
Re: Help finding the local min in a peicewise function

thinkbot said:
f(x) = {
{ 16 - x^2, if -4 <= x <= 0
{ 2x - 3, if 0 <= x <= 4
I did the first derivative test and got zero but i don't think that's right. Any help?
As MarkFL pointed out, you can have f(0)= 16 or -3 but not both.

The derivative of 16- x^2 is -2x which is 0 at x= 0 which is a "break point". Of course, the derivative of 2x- 3 is never 0.

That tells you that a local minimum must occur at x= -4, x= 0, or x= 4. Without knowing if f(0) is 16 or -3, we cannot say which or even if f has a local minimum.
 

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