The formula for linearization L(x) is
\(L(x)=f(a)+f'(x)(x-a)\)
we have
\(f(x)=\sqrt{1-x}\) with \(a=0\)
\(f(a)=\sqrt{1-0}=1\)
\(f'(x)=(\sqrt{1-x})'=(1-x)'(\sqrt{1-x})'=-\frac{1}{2}\frac{1}{\sqrt{1-x}}\)
\(f'(a)=-\frac{1}{2}\frac{1}{\sqrt{1-0}}=-\frac{1}{2}\)
\(L(x)=1-\frac{1}{2}(x-a)=1-\frac{1}{2}x\)
\(\sqrt{1-x}=\sqrt{0.99}\) so \(x=0.01\)
\(L(0.01)=1-\frac{1}{2}0.01=0.95\)
\(\sqrt{1-x}=\sqrt{0.999}\) so \(x=0.001\)
\(L(0.01)=1-\frac{1}{2}0.001=0.995\)
Question
Find the linear approximation of the function f(x)=\sqrt{1-x} at a=0 and use it to approximate the numbers \sqrt{0.9} and \sqrt{0.99}.
Answers (1)
2021-06-12
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