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Plotting a function graph/ 2*log((x+3)/(x-3))-3

Graphing y = 2*log((x+3)/(x-3))-3

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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            /x + 3\    
f(x) = 2*log|-----| - 3
            \x - 3/
f(x)=2log⁡(x+3x−3)−3f{\left(x \right)} = 2 \log{\left(\frac{x + 3}{x - 3} \right)} - 3 
f = 2*log((x + 3)/(x - 3)) - 3
The graph of the function
02468-8-6-4-2-1010-2020
The domain of the function
The points at which the function is not precisely defined:
x1=3x_{1} = 3
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:
2log⁡(x+3x−3)−3=02 \log{\left(\frac{x + 3}{x - 3} \right)} - 3 = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=3tanh⁡(34)x_{1} = \frac{3}{\tanh{\left(\frac{3}{4} \right)}}
Numerical solution
x1=4.72330150073321x_{1} = 4.72330150073321
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 2*log((x + 3)/(x - 3)) - 3.
−3+2log⁡(3−3)-3 + 2 \log{\left(\frac{3}{-3} \right)}
The result:
f(0)=−3+2iπf{\left(0 \right)} = -3 + 2 i \pi
The point:
(0, -3 + 2*pi*i)
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−3)(1x−3−x+3(x−3)2)x+3=0\frac{2 \left(x - 3\right) \left(\frac{1}{x - 3} - \frac{x + 3}{\left(x - 3\right)^{2}}\right)}{x + 3} = 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(1−x+3x−3)(−1x+3−1x−3)x+3=0\frac{2 \left(1 - \frac{x + 3}{x - 3}\right) \left(- \frac{1}{x + 3} - \frac{1}{x - 3}\right)}{x + 3} = 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=3x_{1} = 3

lim⁡x→3−(2(1−x+3x−3)(−1x+3−1x−3)x+3)=∞\lim_{x \to 3^-}\left(\frac{2 \left(1 - \frac{x + 3}{x - 3}\right) \left(- \frac{1}{x + 3} - \frac{1}{x - 3}\right)}{x + 3}\right) = \infty
lim⁡x→3+(2(1−x+3x−3)(−1x+3−1x−3)x+3)=∞\lim_{x \to 3^+}\left(\frac{2 \left(1 - \frac{x + 3}{x - 3}\right) \left(- \frac{1}{x + 3} - \frac{1}{x - 3}\right)}{x + 3}\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=3x_{1} = 3
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(2log⁡(x+3x−3)−3)=−3\lim_{x \to -\infty}\left(2 \log{\left(\frac{x + 3}{x - 3} \right)} - 3\right) = -3
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=−3y = -3
lim⁡x→∞(2log⁡(x+3x−3)−3)=−3\lim_{x \to \infty}\left(2 \log{\left(\frac{x + 3}{x - 3} \right)} - 3\right) = -3
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=−3y = -3
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 2*log((x + 3)/(x - 3)) - 3, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(2log⁡(x+3x−3)−3x)=0\lim_{x \to -\infty}\left(\frac{2 \log{\left(\frac{x + 3}{x - 3} \right)} - 3}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(2log⁡(x+3x−3)−3x)=0\lim_{x \to \infty}\left(\frac{2 \log{\left(\frac{x + 3}{x - 3} \right)} - 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:
2log⁡(x+3x−3)−3=2log⁡(3−x−x−3)−32 \log{\left(\frac{x + 3}{x - 3} \right)} - 3 = 2 \log{\left(\frac{3 - x}{- x - 3} \right)} - 3
- No
2log⁡(x+3x−3)−3=3−2log⁡(3−x−x−3)2 \log{\left(\frac{x + 3}{x - 3} \right)} - 3 = 3 - 2 \log{\left(\frac{3 - x}{- x - 3} \right)}
- No
so, the function
not is
neither even, nor odd
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