Question

How do you find the linearization at a=16 of `f(x) = x^(1/2)`?

Answer 1

Use the formula `L(x)=f(a)+f'(a)(x-a)` to get `L(x)=4+1/8(x-16)=1/8x+2` as the linearization of `f(x)=x^{1/2}` at `a=16`.

Explanation:

For `f(x)=x^{1/2}` we have `f'(x)=1/2 x^{-1/2}` so that `f(a)=f(16)=16^{1/2}=4` and `f'(a)=f'(16)=1/2 * 16^{-1/2}=1/2 * 1/4 = 1/8`.

Therefore, the function `L(x)=f(a)+f'(a)(x-a)=4+1/8(x-16)=1/8x+2` is the linearization of `f(x)=x^{1/2}` at `a=16`.

This can be used to obtain good approximations to square roots of numbers near 16. For example,

`sqrt{16.5}=f(16.5) approx L(16.4)=4+1/8(16.5-16)=4+1/8 * 1/2 = 4+ 1/16=4.0625`.

A more accurate approximation with technology is:

`sqrt{16.5} approx 4.062019202`.

The error in our approximation is about `4.062019202-4.0625 approx -0.00048`, which is pretty good.