##\frac{x - 1}{x + 1} = 1 - \frac{2}{x + 1}##kolua said:
For the vertical asymptote, ask yourself this: can you take the arctan of an undefined value? When will ##\frac{x-1}{x+1}## be undefined?kolua said:
I think the limit of arctan((x-1)/(x+1)) as x approaches -1 from the right is pi/2 and as x approaches -1 from the left is -pi/2, which means there is no vertical asymptote here but a jump discontinuity here right?Mark44 said:
when x=-1 arctan is undefined but as x approaches x=-1 (x-1)/(x+1) approaches negative infinity and positive infinity from left and right. Let (x-1)/(x+1) = t, then arctan(t) approaches pi/2 and -pi/2 as t approaches the two infinity is this correct? then there is no vertical asymptote?Isaac0427 said:
I don't believe that is so. It is undefined at x=-1, but I don't believe that as x approaches -1 that the function approaches infinity with either sign.kolua said:
No, x=-1 is the vertical asymptote.kolua said:
That's not correct.Isaac0427 said:
Ok, I am probably thinking of an incorrect definition of asymptote. The function does get infinitely close to but never reaches π/4 as x approaches -1. Does this not make it an asymptote?Samy_A said:
x approaches infinity and function goes to pi/4 that's the definition of a horizontal asymptote, correct?Isaac0427 said:
Yes that it definitely an asymptote. I'm just not sure about x=-1.kolua said:
No. A function f has a vertical asymptote at x=a if ##\displaystyle\lim_{x\rightarrow a^+}f(x)=\pm \infty## or ##\displaystyle\lim_{x\rightarrow a^-}f(x)=\pm \infty##.Isaac0427 said:
Ok, my bad.Samy_A said:
That's the graph my calculator gives.kolua said:
You misunderstood me. I was answering @Isaac0427 's post concerning vertical asymptotes. The definition of a vertical asymptote makes it evident that your function doesn't have vertical asymptotes. At x=-1 it has a jump discontinuity, exactly as you wrote.kolua said:
The domain of arctan((x-1)/(x+1)) is all real numbers except for -1, as the function is undefined at this point.
To sketch the graph, it is helpful to first find the asymptotes. In this case, the vertical asymptote is x = -1 and the horizontal asymptote is y = π/2. Then, plot a few points on either side of the vertical asymptote to get an idea of the shape of the graph. The end result should be a curve that approaches the asymptotes on either side.
No, the graph is not symmetrical. It is only symmetrical about the point (0,0).
The graph approaches the vertical asymptote from both sides, but never touches it. As x gets closer to -1, the value of the function approaches negative infinity on the left side and positive infinity on the right side.
The value of x affects the value of the function by changing the slope and the vertical shift of the graph. As x approaches infinity, the graph becomes increasingly horizontal. As x approaches -1, the graph becomes increasingly vertical. This is because the value of x in the denominator is causing the function to approach the vertical asymptote.
