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Jusaquikie
 3 years ago
find the linear approximation of f(x)=sqrt 1x at a=0 and use it to approximate sqrt .98
Jusaquikie
 3 years ago
find the linear approximation of f(x)=sqrt 1x at a=0 and use it to approximate sqrt .98

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Jusaquikie
 3 years ago
Best ResponseYou've already chosen the best response.0find the linear approximation of \[f(x)=\sqrt{1x}\] at a=0 and use it to approximate \[\sqrt{.98}\]

Jusaquikie
 3 years ago
Best ResponseYou've already chosen the best response.0i know f(x) = f(a)+f'(a)(xa) and f'(x) here is \[\frac{ 1 }{ 2\sqrt{1x} }\] so \[f'(a)= \frac{ 1 }{\sqrt{10} }=1\] so \[f(x)=\sqrt{10}+(\frac{ 1 }{ 2\sqrt{10} })(x0)\] 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?

Jusaquikie
 3 years ago
Best ResponseYou've already chosen the best response.0\[f'(a)=\frac{ 1 }{2\sqrt{10} }=\frac{ 1 }{ 2 }\]i messed that up when i typed it but put it in the equation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I may be off here... but if linear approx of (1x)^0.5 is f(x) = 11/2*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 (10.02)^0.5?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{1x} \approx 1 \frac{ x }{ 2 }\] Therefore \[\sqrt{0.98} = \sqrt{1  0.02} \approx 1  \frac{ 0.02 }{ 2 }\]

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0agree with him ^ put x=0.02.

Jusaquikie
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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 1x through me and i wasn't sure how to show my work.

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0in general, its good to know that \(\huge \sqrt[n]{1x} \approx 1nx\)

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0\(\huge \sqrt[n]{1 \pm x} \approx 1 \pm nx\)
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