f '(x) = ½(1 - x)^(-½)(-1) = -1/[2√(1 - x)].
f(0) = 1, and f '(0) = -1/2.
So the linear approximation (a.k.a. the tangent line) to f at zero is
L(x) = 1 - x/2.
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f '(x) = ½(1 - x)^(-½)(-1) = -1/[2√(1 - x)].
f(0) = 1, and f '(0) = -1/2.
So the linear approximation (a.k.a. the tangent line) to f at zero is
L(x) = 1 - x/2.