Linear approximation problems. Sanity check

Question

I am a bit new to linear approximation problems and I wanted to check that these were correct.

1. (1.999)^4. Find by linear approximation. Find linearization at $a = 2$. $$f(x) = x^4$$ $$f'(x) = 4x^3$$ so $$L(x) = f(a) + f'(a)(x-a)$$ $$L(x) = 16 + 32(x-2) = 16 + 32x - 64 = -48 + 32x$$

so $(1.999)^4 \approx 16 + 32 ( 1.999-2) = -48 + 32(1.999) = 15.968$

Is this the right process?

What's the right way to think of a? Is the right way to think of it as a starting point for the linearization? It's sort of the base?

Answer 1

Your method is correct and it corresponds to the binomial expansion from which the following linearization holds for $x\to 0$

$$(1\pm x)^a\approx 1\pm xa$$

that is

$$(1.999)^4=(2-0.001)^4=2^4\left(1-\frac{0.001}2\right)^4\approx 16(1-0.002)=15.968$$

Answer 2

Yes, this is the right process. $a$ is the starting point for the linearization. The linearization will be more accurate the closer $x$ is to $a$. The next term is $\frac 12(x-a)^2f''(a)$, which gives you an idea about the error of your linearization.