Question

Use a Linear (or differentials) to estimate square root of 405.

Answer 1

Here's the pertinent equation:

f(x) is approx. equal to f(x0) + f '(x0)(x-x0),
where x0 is the starting point (which would be 400 here, since 400 is close to 405 and is a perfect square).

Then f(405) is approx equal to sqrt(400) + (1/2)(400)^(-1/2) (405-400), or

20 + (1/2)(1/20)(5) = 20.125

The actual square root is sqrt(405) = 20.1246. So we're really close!