LaddiusMaximus
  • LaddiusMaximus
find the linear approximation of the fnct f(x)=sqrt(1-x) at a=0 and use it to approximate the numbers (sqrt0.9) and (sqrt0.99)
Mathematics
chestercat
  • chestercat
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

myininaya
  • myininaya
So do you know how to find the tangent line at x=0?
myininaya
  • myininaya
\[y=mx+b <--\text{ line} \] m=f ' (x=a)=f ' (x=0)=f ' (0) You need to find f ' (x). Then evaluate f ' (x) at x=0. y=f ' (0) x + b We still need to find b though. But we are given a point on this line (0,f(0)) What is f(0)? We will plug in what we know to find what b is: f(0)=f ' (0) * 0 +b f(0)=b SO great.We have the tangent line here at x=0 is y=f ' (0) x + f(0)
myininaya
  • myininaya
Let me know when you find what f ' (0) and f(0) is, then I will give you further assistance.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

LaddiusMaximus
  • LaddiusMaximus
i was showed some different L(x)= f(a)+f'(a)(x-a)

Looking for something else?

Not the answer you are looking for? Search for more explanations.