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 January 13th, 2013, 03:12 AM #1 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 arctan... [color=#0000BF]Dear All! Please, if we have the function: $f(x)=arctan\frac{x+1}{x-1}+arctanx$ then it should be $y=\frac{-1+1}{x^2+1}$ isn't it? then we have the resul 0/x^2+1, isnt it true? but the result says the function is not defined at 1? why? as we habve rational function pol shoudl be x^2+1=0 so x^2=-1 and it shou dnot have result in R shoudl it? the result says the function is constant on $(\infty, 1)$ and [latex]((1, \infty[/latex dont get this...][/color]
 January 13th, 2013, 03:40 AM #2 Member     Joined: Jul 2012 Posts: 60 Thanks: 0 Math Focus: Calculus Re: arctan... $arctg(a) + arctg(b)= arctg\frac{a+b}{1-ab}$ $y=arctg\frac{x+1}{x-1}+arctg(x)$ $=arctg\frac{\frac{x+1}{x-1}+x}{1-\frac{x+1}{x-1}x}$ $=arctg\frac{\frac{x+1+x^2-x}{x-1}}{1-\frac{x^2+x}{x-1}}$ $=arctg\frac{\frac{1+x^2}{x-1}}{\frac{x-1-x^2-x}{x-1}}$ $=arctg\frac{\frac{1+x^2}{x-1}}{\frac{-1-x^2}{x-1}}$ $=arctg\frac{\frac{1+x^2}{x-1}}{\frac{x-1-x^2-x}{x-1}}$ $=arctg\frac{1+x^2}{-(x^2+1)}$ $=arctg(-1)$ "x" can't be 1 because in the initial equation you had "x-1" in the denominator if "x" was 1 the the initial equation would have been: arctg(x+1)/(0) + arctg(x)
 January 13th, 2013, 07:06 AM #3 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 Re: arctan... [color=#0000FF]Hey,sorry, mymistakeagain...it should be the derivative from that function! I got the derivative is y'=0/x^2+1=0 then, for all x<1, the function should be -pi/4,what I dont knowhow to get and for x>1f(x) should be arctan2+arctan3=2.35,what I cant get too... Please,can anyone help? Many thanks![/color]
 January 16th, 2013, 04:17 AM #4 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 Re: arctan... [color=#0000FF]Dear All! Please, how do we get that $f(x)= arctan\frac{x+1}{x-1}+arctanx=arctan2+arctan3?$ f'x(x)=0 f(0=arctan2+arctan3??? for x>1how do I get this? how do I get -pi/4 for x<1??? there is a function: http://www.wolframalpha.com/input/?i=f% ... %2Barctanx Many thanks!!![/color]

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