Find the linear approximation of the function f(x)=\sqrt{4-x} at a=0

Find the linear approximation of the function $f\left(x\right)=\sqrt{4-x}$ at $a=0$ and use it to approximate the numbers $\sqrt{3.9}$ and $\sqrt{3.99}$, please.
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enhebrevz
Use the Taylor Form: $f\left(x\right)=f\left(a\right)+\left(x-a\right)×{f}^{\prime }\left(a\right)\dots$
$f\left(x\right)={\left(\sqrt{4-x}\right)}^{\prime }=\frac{1}{2}×\frac{-1}{\sqrt{4-x}}$
$f\left(0\right)=\sqrt{4-0}=2$
${f}^{\prime }\left(0\right)=\frac{1}{2}×\frac{-1}{\sqrt{4}}=-\frac{1}{4}$
$\sqrt{3.9}=f\left(x\right)\approx 2+0.1\left(-\frac{1}{4}\right)=1.9750$
$\sqrt{3.99}=f\left(x\right)\approx 2+0.01\left(-\frac{1}{4}\right)=1.9975$