Question

need help please on the attachment

Answer 1

find the linear approximation for
\[f(x)=\sqrt{16+x}\] at x = 0

Answer 2

use the point (0,4) and the slope
\[f'(x)=\frac{1}{2\sqrt{16+x}}\] so at 0 the slope is
\[f'(0)=\frac{1}{2\sqrt{16}}=\frac{1}{8}\]

Answer 3

then the equation for the line is
\[y=\frac{1}{8}x+4\]

Answer 4

if
\[16+x=16.2\] then
\[x=.2\] so your approximation is
\[y=\frac{1}{8}\times .02+4\]

Answer 5

as usual a typo it should be
\[y=\frac{1}{8}\times 0.2+4\]

Answer 6

if one of the answers says
\[4\tfrac{1}{40}\] that is the right one

Answer 7

why did you use sqrt(16+x)?