## Jusaquikie 2 years ago find the linear approximation of f(x)=sqrt 1-x at a=0 and use it to approximate sqrt .98

1. Jusaquikie

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

2. Jusaquikie

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?

3. Jusaquikie

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

4. Jusaquikie

5. Xetion

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?

6. Xetion

$\sqrt{1-x} \approx 1- \frac{ x }{ 2 }$ Therefore $\sqrt{0.98} = \sqrt{1 - 0.02} \approx 1 - \frac{ 0.02 }{ 2 }$

7. hartnn

agree with him ^ put x=0.02.

8. Jusaquikie

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.

9. hartnn

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

10. hartnn

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