Help finding the local min in a piecewise function

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SUMMARY

The discussion centers on finding the local minimum of the piecewise function defined as f(x) = {16 - x^2 for -4 <= x <= 0, 2x - 3 for 0 <= x <= 4}. Participants clarify that the critical point at x=0 presents ambiguity due to the function having two values at that point. The first derivative test reveals that the derivative of 16 - x^2 is -2x, which equals zero at x=0, while the derivative of 2x - 3 does not yield any critical points. Consequently, local minima can only occur at the endpoints x = -4, x = 0, or x = 4, but the exact minimum cannot be determined without resolving the value of f(0).

PREREQUISITES
  • Understanding of piecewise functions
  • Knowledge of first derivative tests in calculus
  • Familiarity with critical points and their significance
  • Ability to analyze function behavior at breakpoints
NEXT STEPS
  • Study the properties of piecewise functions in depth
  • Learn about critical points and local extrema in calculus
  • Explore the implications of breakpoints in piecewise-defined functions
  • Investigate the concept of continuity and limits at points of discontinuity
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Students and educators in calculus, mathematicians analyzing piecewise functions, and anyone seeking to understand local minima in mathematical functions.

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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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