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OpenStudy (anonymous):

what is the square root of 129 when using linear approximation

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OpenStudy (anonymous):

this is calc 1 right?

OpenStudy (anonymous):

yeah it is claculus

OpenStudy (anonymous):

calculus*

OpenStudy (anonymous):

i do not get how you got the second part

OpenStudy (anonymous):

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

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OpenStudy (anonymous):

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

OpenStudy (anonymous):

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 \(y=\frac{1}{14}x+7\)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thank you satellite73 (:

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