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Plotting a function graph/ arctg((x-1)/(x+1))

Graphing y = arctg((x-1)/(x+1))

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
           /x - 1\
f(x) = atan|-----|
           \x + 1/
f(x)=atan⁡(x−1x+1)f{\left(x \right)} = \operatorname{atan}{\left(\frac{x - 1}{x + 1} \right)} 
f = atan((x - 1)/(x + 1))
The graph of the function
02468-8-6-4-2-10105-5
The domain of the function
The points at which the function is not precisely defined:
x1=−1x_{1} = -1
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis X at f = 0
so we need to solve the equation:
atan⁡(x−1x+1)=0\operatorname{atan}{\left(\frac{x - 1}{x + 1} \right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=1x_{1} = 1
Numerical solution
x1=1x_{1} = 1
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to atan((x - 1)/(x + 1)).
atan⁡(−1−1)\operatorname{atan}{\left(- 1^{-1} \right)}
The result:
f(0)=−π4f{\left(0 \right)} = - \frac{\pi}{4}
The point:
(0, -pi/4)
Extrema of the function
In order to find the extrema, we need to solve the equation
ddxf(x)=0\frac{d}{d x} f{\left(x \right)} = 0
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
the first derivative
−x−1(x+1)2+1x+1(x−1)2(x+1)2+1=0\frac{- \frac{x - 1}{\left(x + 1\right)^{2}} + \frac{1}{x + 1}}{\frac{\left(x - 1\right)^{2}}{\left(x + 1\right)^{2}} + 1} = 0
Solve this equation
Solutions are not found,
function may have no extrema
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
d2dx2f(x)=0\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
2(x−1x+1−1)(−(x−1)(x−1x+1−1)(x+1)((x−1)2(x+1)2+1)+1)(x+1)2((x−1)2(x+1)2+1)=0\frac{2 \left(\frac{x - 1}{x + 1} - 1\right) \left(- \frac{\left(x - 1\right) \left(\frac{x - 1}{x + 1} - 1\right)}{\left(x + 1\right) \left(\frac{\left(x - 1\right)^{2}}{\left(x + 1\right)^{2}} + 1\right)} + 1\right)}{\left(x + 1\right)^{2} \left(\frac{\left(x - 1\right)^{2}}{\left(x + 1\right)^{2}} + 1\right)} = 0
Solve this equation
The roots of this equation
x1=0x_{1} = 0
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
x1=−1x_{1} = -1

lim⁡x→−1−(2(x−1x+1−1)(−(x−1)(x−1x+1−1)(x+1)((x−1)2(x+1)2+1)+1)(x+1)2((x−1)2(x+1)2+1))=0.5\lim_{x \to -1^-}\left(\frac{2 \left(\frac{x - 1}{x + 1} - 1\right) \left(- \frac{\left(x - 1\right) \left(\frac{x - 1}{x + 1} - 1\right)}{\left(x + 1\right) \left(\frac{\left(x - 1\right)^{2}}{\left(x + 1\right)^{2}} + 1\right)} + 1\right)}{\left(x + 1\right)^{2} \left(\frac{\left(x - 1\right)^{2}}{\left(x + 1\right)^{2}} + 1\right)}\right) = 0.5
lim⁡x→−1+(2(x−1x+1−1)(−(x−1)(x−1x+1−1)(x+1)((x−1)2(x+1)2+1)+1)(x+1)2((x−1)2(x+1)2+1))=0.5\lim_{x \to -1^+}\left(\frac{2 \left(\frac{x - 1}{x + 1} - 1\right) \left(- \frac{\left(x - 1\right) \left(\frac{x - 1}{x + 1} - 1\right)}{\left(x + 1\right) \left(\frac{\left(x - 1\right)^{2}}{\left(x + 1\right)^{2}} + 1\right)} + 1\right)}{\left(x + 1\right)^{2} \left(\frac{\left(x - 1\right)^{2}}{\left(x + 1\right)^{2}} + 1\right)}\right) = 0.5
- limits are equal, then skip the corresponding point

Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
(−∞,0]\left(-\infty, 0\right]
Convex at the intervals
[0,∞)\left[0, \infty\right)
Vertical asymptotes
Have:
x1=−1x_{1} = -1
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞atan⁡(x−1x+1)=π4\lim_{x \to -\infty} \operatorname{atan}{\left(\frac{x - 1}{x + 1} \right)} = \frac{\pi}{4}
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=π4y = \frac{\pi}{4}
lim⁡x→∞atan⁡(x−1x+1)=π4\lim_{x \to \infty} \operatorname{atan}{\left(\frac{x - 1}{x + 1} \right)} = \frac{\pi}{4}
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=π4y = \frac{\pi}{4}
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of atan((x - 1)/(x + 1)), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(atan⁡(x−1x+1)x)=0\lim_{x \to -\infty}\left(\frac{\operatorname{atan}{\left(\frac{x - 1}{x + 1} \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(atan⁡(x−1x+1)x)=0\lim_{x \to \infty}\left(\frac{\operatorname{atan}{\left(\frac{x - 1}{x + 1} \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) и f = -f(-x).
So, check:
atan⁡(x−1x+1)=atan⁡(−x−11−x)\operatorname{atan}{\left(\frac{x - 1}{x + 1} \right)} = \operatorname{atan}{\left(\frac{- x - 1}{1 - x} \right)}
- No
atan⁡(x−1x+1)=−atan⁡(−x−11−x)\operatorname{atan}{\left(\frac{x - 1}{x + 1} \right)} = - \operatorname{atan}{\left(\frac{- x - 1}{1 - x} \right)}
- No
so, the function
not is
neither even, nor odd
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