When using linear approximation to estimate 1/4.002, what is the x in L(x) = f(a)-f'(a)(x-a)

Question

Sorry in advance for the formatting I've never used this site before.

I'm asked to use linear approximation to estimate 1/4.002

I found the derivative = -1/(x^2)

and I believe f(x) is 1/4

so a = 4

and L(x) = f(a) - f'(a)(x-a)

therefore: L(x) = 1/4 - (-1/16)(x-4)

But what is the x? I'm thinking its 0.002 but that doesn't seem right cause it need to be close to 4. Is it just 4.002?

Answer 1

Yes. It is just 4.002. You are estimating $f(x)=1/x$ at $x=4.002$ by evaluating $L(x)$ there instead.

Answer 2

See, the first degree approximation to $f(x)$ is given by $f(a) + f'(a)(x-a)$.

You want the value of $f(4.002)$, where $f(x) = \frac 1x$. $a$ is a value close to $x$ such that $f(a)$ is easy to calculate. So in your case, $a=4$ is desirable.

Now, if you plug in the formula with $a = 4$ and $x = 4.002$, you get: $$ f(4) + f'(4)(4.002-4) = \frac 14 - \frac{0.002}{16} = 0.249875. $$

This is the first degree approximation to $\frac 1{4.002}$. The actual value is somewhat like $0.249875062...$ so you are actually correct to six decimal places for even a first degree approximation.