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