oliviayychengwh

2021-12-18

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.

enhebrevz

Beginner2021-12-19Added 25 answers

Use the Taylor Form: $f\left(x\right)=f\left(a\right)+(x-a)\times {f}^{\prime}\left(a\right)\dots$

$f\left(x\right)={\left(\sqrt{4-x}\right)}^{\prime}=\frac{1}{2}\times \frac{-1}{\sqrt{4-x}}$

$f\left(0\right)=\sqrt{4-0}=2$

$f}^{\prime}\left(0\right)=\frac{1}{2}\times \frac{-1}{\sqrt{4}}=-\frac{1}{4$

$\sqrt{3.9}=f\left(x\right)\approx 2+0.1(-\frac{1}{4})=1.9750$

$\sqrt{3.99}=f\left(x\right)\approx 2+0.01(-\frac{1}{4})=1.9975$