Question

square root of 398 solving calculus way

Answer 1

go to calculatersoup.com search square roots

Answer 2

what does it mean "calculus way"? maybe finding the linearization of
\[f(x)=\sqrt{400+x}\] at x = 0?

Answer 3

if so find
\[f(0)=20\]
\[f'(x)=\frac{1}{2\sqrt{400+x}}\]
\[f'(0)=\frac{1}{40}\] and so the line is
\[y=\frac{1}{40}x+20\]

Answer 4

then for 389 put
\[400+x=389\]
\[x=-2\] and so your linear approximation is
\[y=\frac{-2}{40}+20\] or
\[-.05+20=19.95\]