Question

what is the square root of 129 when using linear approximation

Answer 1

this is calc 1 right?

Answer 2

yeah it is claculus

Answer 3

calculus*

Answer 4

i do not get how you got the second part

Answer 5

you can also try the linear approximation to \(\sqrt{49+x}\) near zero, then replace \(x\) by 6
it is a rather lousy approximation though

Answer 6

y(x+x1)=m(x) (x-x1) + y(x)

Answer 7

i will be interested if we get the same answer
the derivative of \(\sqrt{49+x}\) is \(\frac{1}{2\sqrt{49+x}}\) at zero you get the slope as \(\frac{1}{14}\) then using the point slope formula at the point \((0,7)\) the equation for the line is \(y=\frac{1}{14}x+7\)

Answer 8

then if you replace \(x\) by \(6\) you get the lousy approximation
\[\frac{6}{14}+7\]

Answer 9

thank you satellite73 (: