Question

Let be the function defined by $f(x)=\sqrt{x}$. What is the approximation for the value of $\sqrt{3}$ obtained by using the second-degree Taylor polynomial for $f$ about $x=4$?

Answer 1

Answer:

Step-by-step explanation:

The Taylor series at x=4 is

$\sqrt{x} = 2 + \frac{x-4}{4} - \frac{(x-4)^2}{64} + ...$

So,

$\sqrt{3} = 2 - \frac{1}{4} - \frac{1}{64} = 1.73$

Not too bad, eh?