This function is defined on [-1, 1] and is differentiable on (-1, 1). Its derivative is f'(x) = 1/sqrt(1-x^2) - 2.
The derivative doesn't exist at x = +-1. It is zero where 1-x^2 = 1/4, so at x = +-sqrt(3)/2. It is an even function and it is obviously increases for positive x and decreases for negative x. Hence it is positive on (-1, -sqrt(3)/2) uu (sqrt(3)/2, 1) and negative on (-sqrt(3)/2, sqrt(3)/2), and the function f increases and decreases respectively.
This way we can determine the maximum and minimum of f: -1 is a local (one-sided) minimum, 1 is a local one-sided maximum, -sqrt(3)/2 is the local maximum and sqrt(3)/2 is a local minimum.
Monday, March 21, 2016
Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 70
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