Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

@hartnn Help please
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 \(\huge \sqrt[n]{1-x} \approx 1-nx\)
\(\huge \sqrt[n]{1 \pm x} \approx 1 \pm nx\)

Not the answer you are looking for?

Search for more explanations.

Ask your own question