Here's the question you clicked on:
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)
So do you know how to find the tangent line at x=0?
\[y=mx+b <--\text{ line} \] 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)
Let me know when you find what f ' (0) and f(0) is, then I will give you further assistance.
i was showed some different L(x)= f(a)+f'(a)(x-a)