Local Extremum of f(x)=(4-x^2)^(-1/2): First Derivative Test

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SUMMARY

The discussion centers on the function f(x)=(4-x^2)^(-1/2) and its critical point at x=0. According to the First Derivative Test, if the derivative f' does not change sign at a critical point, then there is no local extremum at that point. However, the book asserts that there is a minimum value at x=0. Participants in the discussion seek clarification on the calculations that demonstrate the behavior of the derivative at this critical point.

PREREQUISITES
  • Understanding of calculus concepts, specifically the First Derivative Test.
  • Familiarity with critical points and local extrema.
  • Knowledge of how to compute derivatives of functions.
  • Ability to analyze the behavior of functions around critical points.
NEXT STEPS
  • Review the First Derivative Test in detail.
  • Practice calculating derivatives for various functions, including f(x)=(4-x^2)^(-1/2).
  • Explore the concept of local extrema and how they are determined.
  • Investigate the implications of non-changing sign derivatives at critical points.
USEFUL FOR

Students studying calculus, educators teaching derivative concepts, and anyone interested in understanding local extrema and their determination through the First Derivative Test.

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1. According to the First Derivative Test for local extrema, if f' doesn't change sign at c, then f has no local extreme value at c. But for a question on my book, f(x)=(4-x^2)^(-1/2), the critical point is 0, but i think it doesn't have local extreme because the derivative doesn't change sign ,but my book says it has minimum value at x=0.
Thanks for help.
 
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Can you write out the calculations leading you to conclude that the derivative doesn't change sign at x = 0?
 

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