Verify the given linear approximation at
a = 0.
Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter your answer using interval notation. Round your answers to three decimal places.)
ln(1 + x) ≈ x

Question

Verify the given linear approximation at 
a = 0.
 Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter your answer using interval notation. Round your answers to three decimal places.)
ln(1 + x) ≈ x

anonymous · Answer

I have no idea how to do this one. I think I need to use this formula: L(x)=f(a)+f'(a) (x-a)

anonymous · Answer

it is the same as a problem you probably did in the second or third week of class
find the equation of the line tangent to the curve of $y=f(x)$ at $(a,f(a))$

anonymous · Answer

so...would it be find the derivative of ln(1+x)=x??

anonymous · Answer

oh no

anonymous · Answer

they are giving you the answer

anonymous · Answer

they are telling you that the equation of the line tangent to $y=\ln(1+x)$ at $(0,1)$ is $y=x$ and asking you to "check" that it is right

anonymous · Answer

okay that was wrong, it is at $(0,0)$

anonymous · Answer

so, should I just plug in numbers 0.1 greater and smaller into the formula? I really have no idea what to do

anonymous · Answer

in any case the check is to take the derivative of $\ln(x+1)$ get $\frac{1}{x+1}$ then replace $x$ by $0$ to find the slope, you get that the slope is 1

anonymous · Answer

ok..

anonymous · Answer

and since the line goes through the origin, the equation of the tangent line at $(0,0)$ is $y=x$

anonymous · Answer

wait! so would the answer be $$x \in \mathbb{R} $$  ??

anonymous · Answer

no the linear approximation is only good locally, not globally

anonymous · Answer

I'm sorry, I don't know what to do next, I understand everything else you've done but I dont know how to get what they are asking for

anonymous · Answer

okay first here is a picture, then we can see how for what values of x we get a small error http://www.wolframalpha.com/input/?i=ln%28x%2B1%29%2C+x+++domain+-0.2..0.2

anonymous · Answer

without using power series i am not sure exactly how to answer the last question
you want to make $|\ln(x+1)-x|<0.1$
 or $\ln(x+1)-.1<x<\ln(x+1)+.1$

anonymous · Answer

so i guess you can just check with numbers and see what works

anonymous · Answer

any number?? or shouls I use small decimals?

anonymous · Answer

small ones
i don't think this is that easy

anonymous · Answer

ok, thanks!

anonymous · Answer

yw, sorry i couldn't be more help