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

Question

23428 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-25T21:07:06

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 $a=16$ .

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.

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