# How do I approximate sqrt(128) using a Taylor polynomial centered at 125?

##### 1 Answer
Mar 8, 2015

To approximating $f \left(x\right) = \sqrt{x}$ by a Taylor polynomial centered at $125$ , you'll need to know $\sqrt{125}$.

The Taylor polynomial, of degree $n + 1$, to approximate $f \left(x\right)$ centered at $a$ involves:

$f \left(a\right) , f ' \left(a\right) , f ' ' \left(a\right) , f ' ' ' \left(a\right) , . . . , {f}^{\left(n + 1\right)} \left(a\right)$

and also involves $x - a$

Derivatives of $\sqrt{x}$ of every order will involve ${\sqrt{x}}^{k}$, Thus, every term will require $\sqrt{125}$

Re-read the question you were asked.

The exercise would make more sense if either:
(a) the question is to approximate $\sqrt[3]{128}$
($f \left(x\right) = \sqrt[3]{x}$ and its derivatives are not too hard to find when $x = 125$

or

(b) $a$ is not specified
(In which case, note that $\sqrt{128} = \sqrt{{2}^{7}} = {2}^{\frac{7}{2}}$ so you can use $f \left(x\right) = {x}^{\frac{7}{2}}$ centered at $1$ to approximate $f \left(2\right)$)