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Plotting a function graph/ 1/3+ln((2+x)/(2-x))

Graphing y = 1/3+ln((2+x)/(2-x))

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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       1      /2 + x\
f(x) = - + log|-----|
       3      \2 - x/
f(x)=log⁡(x+22−x)+13f{\left(x \right)} = \log{\left(\frac{x + 2}{2 - x} \right)} + \frac{1}{3} 
f = log((x + 2)/(2 - x)) + 1/3
The graph of the function
02468-8-6-4-2-1010-1010
The domain of the function
The points at which the function is not precisely defined:
x1=2x_{1} = 2
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:
log⁡(x+22−x)+13=0\log{\left(\frac{x + 2}{2 - x} \right)} + \frac{1}{3} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−2tanh⁡(16)x_{1} = - 2 \tanh{\left(\frac{1}{6} \right)}
Numerical solution
x1=−0.330280825849259x_{1} = -0.330280825849259
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 1/3 + log((2 + x)/(2 - x)).
log⁡(22−0)+13\log{\left(\frac{2}{2 - 0} \right)} + \frac{1}{3}
The result:
f(0)=13f{\left(0 \right)} = \frac{1}{3}
The point:
(0, 1/3)
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
(2−x)(12−x+x+2(2−x)2)x+2=0\frac{\left(2 - x\right) \left(\frac{1}{2 - x} + \frac{x + 2}{\left(2 - x\right)^{2}}\right)}{x + 2} = 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
(1−x+2x−2)(−1x+2−1x−2)x+2=0\frac{\left(1 - \frac{x + 2}{x - 2}\right) \left(- \frac{1}{x + 2} - \frac{1}{x - 2}\right)}{x + 2} = 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=2x_{1} = 2

lim⁡x→2−((1−x+2x−2)(−1x+2−1x−2)x+2)=∞\lim_{x \to 2^-}\left(\frac{\left(1 - \frac{x + 2}{x - 2}\right) \left(- \frac{1}{x + 2} - \frac{1}{x - 2}\right)}{x + 2}\right) = \infty
lim⁡x→2+((1−x+2x−2)(−1x+2−1x−2)x+2)=∞\lim_{x \to 2^+}\left(\frac{\left(1 - \frac{x + 2}{x - 2}\right) \left(- \frac{1}{x + 2} - \frac{1}{x - 2}\right)}{x + 2}\right) = \infty
- 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[0, \infty\right)
Convex at the intervals
(−∞,0]\left(-\infty, 0\right]
Vertical asymptotes
Have:
x1=2x_{1} = 2
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(log⁡(x+22−x)+13)=13+iπ\lim_{x \to -\infty}\left(\log{\left(\frac{x + 2}{2 - x} \right)} + \frac{1}{3}\right) = \frac{1}{3} + i \pi
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=13+iπy = \frac{1}{3} + i \pi
lim⁡x→∞(log⁡(x+22−x)+13)=13+iπ\lim_{x \to \infty}\left(\log{\left(\frac{x + 2}{2 - x} \right)} + \frac{1}{3}\right) = \frac{1}{3} + i \pi
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=13+iπy = \frac{1}{3} + i \pi
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 1/3 + log((2 + x)/(2 - x)), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(log⁡(x+22−x)+13x)=0\lim_{x \to -\infty}\left(\frac{\log{\left(\frac{x + 2}{2 - x} \right)} + \frac{1}{3}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(log⁡(x+22−x)+13x)=0\lim_{x \to \infty}\left(\frac{\log{\left(\frac{x + 2}{2 - x} \right)} + \frac{1}{3}}{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:
log⁡(x+22−x)+13=log⁡(2−xx+2)+13\log{\left(\frac{x + 2}{2 - x} \right)} + \frac{1}{3} = \log{\left(\frac{2 - x}{x + 2} \right)} + \frac{1}{3}
- No
log⁡(x+22−x)+13=−log⁡(2−xx+2)−13\log{\left(\frac{x + 2}{2 - x} \right)} + \frac{1}{3} = - \log{\left(\frac{2 - x}{x + 2} \right)} - \frac{1}{3}
- No
so, the function
not is
neither even, nor odd
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