Question

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

Answer 1

find the linear approximation of
\[f(x)=\sqrt{1-x}\] at a=0
and use it to approximate \[\sqrt{.98}\]

Answer 2

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?

Answer 3

\[f'(a)=\frac{ -1 }{2\sqrt{1-0} }=-\frac{ 1 }{ 2 }\]i messed that up when i typed it but put it in the equation

Answer 4

@hartnn Help please

Answer 5

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?

Answer 6

\[\sqrt{1-x} \approx 1- \frac{ x }{ 2 }\]

Therefore \[\sqrt{0.98} = \sqrt{1 - 0.02} \approx 1 - \frac{ 0.02 }{ 2 }\]

Answer 7

agree with him ^
put x=0.02.

Answer 8

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.

Answer 9

in general, its good to know that
\(\huge \sqrt[n]{1-x} \approx 1-nx\)

Answer 10

\(\huge \sqrt[n]{1 \pm x} \approx 1 \pm nx\)