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Jusaquikie
find the linear approximation of f(x)=sqrt 1-x at a=0 and use it to approximate sqrt .98
find the linear approximation of \[f(x)=\sqrt{1-x}\] at a=0 and use it to approximate \[\sqrt{.98}\]
i know f(x) = f(a)+f'(a)(x-a) and f'(x) here is \[\frac{ -1 }{ 2\sqrt{1-x} }\] so \[f'(a)= \frac{ -1 }{\sqrt{1-0} }=-1\] so \[f(x)=\sqrt{1-0}+(\frac{ -1 }{ 2\sqrt{1-0} })(x-0)\] so \[f(x)=1+(-\frac{ 1 }{2 })x-(-\frac{ 1 }{2 })0 = 1-\frac{ 1 }{2 }x\] \[f(.98)= 1-\frac{ 1 }{ 2 }(.98)\] and i'm lost, so where did i go wrong and what should i do?
\[f'(a)=\frac{ -1 }{2\sqrt{1-0} }=-\frac{ 1 }{ 2 }\]i messed that up when i typed it but put it in the equation
I may be off here... but if linear approx of (1-x)^0.5 is f(x) = 1-1/2*x Would you want to plug in 0.02 into the equation not 0.98. Because to get the square root of 0.98 is the same a (1-0.02)^0.5?
\[\sqrt{1-x} \approx 1- \frac{ x }{ 2 }\] Therefore \[\sqrt{0.98} = \sqrt{1 - 0.02} \approx 1 - \frac{ 0.02 }{ 2 }\]
agree with him ^ put x=0.02.
ok thanks i was confused because i was used to approximating using just square root of x and plugging in a value and taking the tangent line etc but the 1-x through me and i wasn't sure how to show my work.
in general, its good to know that \(\huge \sqrt[n]{1-x} \approx 1-nx\)
\(\huge \sqrt[n]{1 \pm x} \approx 1 \pm nx\)