Linear approximation and percentage error

[ver_mathstats]

Homework Statement

Approximate 1/99 to four decimal places using the linearization L(x) of f(x)=1/x at a=100 and use a calculator to compute the percentage error.

Relevant Equations

L(x) of f(x)=1/x

I found the linearization, L(x) = -0.0001x+0.2 and I found L(1/99) = 0.0199989899.

Then I tried to put that value into my percentage error formula along with 1/99 and got:

the absolute value of (1/99)-L(1/99) and then we divide that by our actual value which is 1/99, then I multiply everything by 100.

I got the answer 97.99% and I got this wrong, but I am unsure of where I went wrong and how to fix my mistake.

Thank you.

 

[Mark44]

I found the linearization, L(x) = -0.0001x+0.2 and I found L(1/99) = 0.0199989899.

Can you show your work in getting this? Your value for L(1/99) seems off by quite a bit. $1/99 \approx .01$ and your answer is almost twice this.

the absolute value of (1/99)-L(1/99) and then we divide that by our actual value which is 1/99, then I multiply everything by 100.

That's the idea, but since you didn't show your figures, it's possible that some of them are incorrect.

 

[Orodruin]

Homework Statement: Approximate 1/99 to four decimal places using the linearization L(x) of f(x)=1/x at a=100 and use a calculator to compute the percentage error.
Homework Equations: L(x) of f(x)=1/x

I found the linearization, L(x) = -0.0001x+0.2 and I found L(1/99) = 0.0199989899.

A good check to make for a linearization is that it gives you back the correct value at the point you are linearlizing around.

 

[WWGD]

Where did you get the 0.2 in L(x)= -0.0001x+0.2?