# How do you find the second-degree Taylor polynomial T_2(x) for the function f(x) = sqrt(3+x^2) at the number x=1?

Mar 13, 2017

${T}_{2} \left(x\right) = 2 + \frac{1}{2} \left(x - 1\right) + \frac{3}{16} {\left(x - 1\right)}^{2}$

#### Explanation:

A complete Taylor polynomial for the function $f$ centered around $x = c$ is given by:

T(x)=sum_(n=0)^oo(f^((n))(c))/(n!)(x-c)^n

So the second degree Taylor polynomial will be the sum of the terms $n = 0$ through $n = 2$, or:

T_2(x)=(f^((0))(c))/(0!)(x-c)^0+(f^((1))(c))/(1!)(x-c)^1+(f^((2))(c))/(2!)(x-c)^2" "

$c = 1$ and we can make a couple other simplifications:

${T}_{2} \left(x\right) = f \left(1\right) + f ' \left(1\right) \left(x - 1\right) + \frac{f ' ' \left(1\right)}{2} {\left(x - 1\right)}^{2} \text{ "" } \textcolor{red}{\star}$

We see that:

$f \left(1\right) = \sqrt{3 + 1} = 2$

And:

$f ' \left(x\right) = \frac{1}{2} {\left(3 + {x}^{2}\right)}^{- \frac{1}{2}} \left(2 x\right) = \frac{x}{\sqrt{3 + {x}^{2}}}$

$f ' \left(1\right) = \frac{1}{\sqrt{3 + 1}} = \frac{1}{2}$

Finding $f ' '$ using $f ' \left(x\right) = x {\left(3 + {x}^{2}\right)}^{- \frac{1}{2}}$:

$f ' ' \left(x\right) = {\left(3 + {x}^{2}\right)}^{- \frac{1}{2}} + x \left(- \frac{1}{2} {\left(3 + {x}^{2}\right)}^{- \frac{3}{2}}\right) \left(2 x\right)$

$f ' ' \left(x\right) = {\left(3 + {x}^{2}\right)}^{- \frac{1}{2}} - {x}^{2} {\left(3 + {x}^{2}\right)}^{- \frac{3}{2}}$

$f ' ' \left(x\right) = \frac{3}{3 + {x}^{2}} ^ \left(\frac{3}{2}\right)$

$f ' ' \left(1\right) = \frac{3}{1 + 3} ^ \left(\frac{3}{2}\right) = \frac{3}{8}$

Returning to $\textcolor{red}{\star}$:

${T}_{2} \left(x\right) = 2 + \frac{1}{2} \left(x - 1\right) + \frac{3}{16} {\left(x - 1\right)}^{2}$