# 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}.

Find the linear approximation of the function $f\left(x\right)=\sqrt{1-x}$ at $a=0$ and use it to approximate the numbers $\sqrt{0.9}$ and $\sqrt{0.99}$.

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wheezym

The formula for linearization L(x) is
$L\left(x\right)=f\left(a\right)+{f}^{\prime }\left(x\right)\left(x-a\right)$
we have
$f\left(x\right)=\sqrt{1-x}$ with $a=0$
$f\left(a\right)=\sqrt{1-0}=1$
${f}^{\prime }\left(x\right)=\left(\sqrt{1-x}{\right)}^{\prime }=\left(1-x{\right)}^{\prime }\left(\sqrt{1-x}{\right)}^{\prime }=-\frac{1}{2}\frac{1}{\sqrt{1-x}}$
${f}^{\prime }\left(a\right)=-\frac{1}{2}\frac{1}{\sqrt{1-0}}=-\frac{1}{2}$
$L\left(x\right)=1-\frac{1}{2}\left(x-a\right)=1-\frac{1}{2}x$
$\sqrt{1-x}=\sqrt{0.99}$ so $x=0.01$
$L\left(0.01\right)=1-\frac{1}{2}0.01=0.95$
$\sqrt{1-x}=\sqrt{0.999}$ so $x=0.001$
$L\left(0.01\right)=1-\frac{1}{2}0.001=0.995$