Give the Taylor expansion order 1 for the function:

http://www.mathramz.com/xyz/latexren...2a7211c1bf.png

and dudect that graph offaccept asymtot of which theyask

his psition andgraph off .

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- Jun 10th 2009, 12:12 AMdhiabTaylor expansion
**Give the Taylor expansion order 1 for the function**:

http://www.mathramz.com/xyz/latexren...2a7211c1bf.png

**and dudect that graph of**f**accept asymtot of which they**ask

**his psition and****graph of**f . - Jun 10th 2009, 07:05 AMTheEmptySet
$\displaystyle f(a)=a^2\arctan\left( \frac{1}{a+1}\right) $

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

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

$\displaystyle f'(a)=2a\arctan\left( \frac{1}{a+1}\right)-\left(\frac{2a}{(1+a)^2+1} \right) $

Since you didn't say where it was centered I centered it at "a"

$\displaystyle T_1(x)=f(a)+f'(a)(x-a)$

$\displaystyle T_1(x)=a^2\arctan\left( \frac{1}{a+1}\right)+\left[2a\arctan\left( \frac{1}{a+1}\right)-\left(\frac{2a}{(1+a)^2+1} \right) \right](x-a) $