Verify the given linear approximation at $a=0$. Then determine the values of x for which the linear approximation is accurate within 0.1: $$ln(1 + x) ≈ x $$
The official method that was used to calculate a question similar to this was to use this formula:
$$ |f(x) - L(x)| < 0.1 $$
for example, in this case, it would be:
$$|ln(1 + x) - x| < 0.1 $$
My first question is, how do you solve for x algebraically in the equation above?
My second question is why isn't this the equation?
$$ |f'(x) - L(x)| < 0.1 $$
Logically I think the equation above makes more sense because you want to compare the approximation of the tangent line (L(x)) with the actual tangent line (f'x).
I believe the inequality you would be solving would be
$$|x-\ln(1+x)|<0.1$$
because you're trying to find the difference between your approximation $x$ and the function $\ln(1+x)$ itself.
WolframAlpha gives the following solution to six decimal places:
$$-0.383183<x<0.516221$$