Linear approximation to find $\frac{1}{4.002}$

Question

I am asked to find $\frac{1}{4.002}$ using linear approximation.

The way I proceeded was:

$$ f(x) = \frac{1}{x} \quad a = 4 \quad f(a) = f(4) = 0.25\\ f'(x) = - \frac{1}{x^2} \quad f'(a) = f'(4) = - \frac{1}{16} $$

So

$\frac{1}{4.002} \approx \frac{1}{4} - \frac{1}{16} (4.002 - 4) = 0.249875$

Is this correct?

If I want to find out how good was my approximation to the real value, what should I do?

Thank you.

Answer 1

If $x>0$ is close to zero $$\frac{1}{4+x}=\frac{1}{4}\cdot\frac{1}{1+\frac{x}{4}}=\frac{1}{4}\sum_{n\geq 0}\frac{(-1)^n x^n}{4^n}$$ holds, and since the terms of the last series are decreasing in absolute value, $$ \frac{1}{4}\left(1-\frac{x}{4}\right) \leq \frac{1}{4+x}\leq \frac{1}{4}\left(1-\frac{x}{4}+\frac{x^2}{16}\right) $$ holds, too. Just plug in $x=\frac{1}{500}$ to get an upper bound for the approximation error by the RHS inequality.