f(100) = .01 ; f'(x) = - 1/x2 ; f'(100) = - .0001
L(98) = .01 - .0001(-2) = .0102
f(98) = .0102041863
[f(98) - L(98)] / f(98) = .0004 = .04 %
M P.
asked 06/22/21Approximate1/98 to four decimal places using the linearization L(x) of f(x) = 1/x at a = 100 and use a calculator to compute the percentage error.
(Use decimal notation. Give your answer to two decimal places.)
The percentage error ≈
f(100) = .01 ; f'(x) = - 1/x2 ; f'(100) = - .0001
L(98) = .01 - .0001(-2) = .0102
f(98) = .0102041863
[f(98) - L(98)] / f(98) = .0004 = .04 %
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