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

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

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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          /    3    \
f(x) = log|x - - + 3|
          \    x    /
f(x)=log⁡((x−3x)+3)f{\left(x \right)} = \log{\left(\left(x - \frac{3}{x}\right) + 3 \right)} 
f = log(x - 3/x + 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=0x_{1} = 0
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−3x)+3)=0\log{\left(\left(x - \frac{3}{x}\right) + 3 \right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−3x_{1} = -3
x2=1x_{2} = 1
Numerical solution
x1=−3x_{1} = -3
x2=1x_{2} = 1
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to log(x - 3/x + 3).
log⁡(3−30)\log{\left(3 - \frac{3}{0} \right)}
The result:
f(0)=∞~f{\left(0 \right)} = \tilde{\infty}
sof doesn't intersect Y
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
1+3x2(x−3x)+3=0\frac{1 + \frac{3}{x^{2}}}{\left(x - \frac{3}{x}\right) + 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
−(1+3x2)2x+3−3x+6x3x+3−3x=0- \frac{\frac{\left(1 + \frac{3}{x^{2}}\right)^{2}}{x + 3 - \frac{3}{x}} + \frac{6}{x^{3}}}{x + 3 - \frac{3}{x}} = 0
Solve this equation
The roots of this equation
x1=−−8+2338014+18543+2338014+185432+−16−2338014+18543−2338014+18543+36−8+2338014+18543+2338014+185432x_{1} = - \frac{\sqrt{-8 + \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}}}{2} + \frac{\sqrt{-16 - 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}} - \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + \frac{36}{\sqrt{-8 + \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}}}}}{2}
x2=−−16−2338014+18543−2338014+18543+36−8+2338014+18543+2338014+185432−−8+2338014+18543+2338014+185432x_{2} = - \frac{\sqrt{-16 - 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}} - \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + \frac{36}{\sqrt{-8 + \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}}}}}{2} - \frac{\sqrt{-8 + \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}}}{2}
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=0x_{1} = 0

lim⁡x→0−(−(1+3x2)2x+3−3x+6x3x+3−3x)=∞\lim_{x \to 0^-}\left(- \frac{\frac{\left(1 + \frac{3}{x^{2}}\right)^{2}}{x + 3 - \frac{3}{x}} + \frac{6}{x^{3}}}{x + 3 - \frac{3}{x}}\right) = \infty
lim⁡x→0+(−(1+3x2)2x+3−3x+6x3x+3−3x)=∞\lim_{x \to 0^+}\left(- \frac{\frac{\left(1 + \frac{3}{x^{2}}\right)^{2}}{x + 3 - \frac{3}{x}} + \frac{6}{x^{3}}}{x + 3 - \frac{3}{x}}\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
[−−16−2338014+18543−2338014+18543+36−8+2338014+18543+2338014+185432−−8+2338014+18543+2338014+185432,−−8+2338014+18543+2338014+185432+−16−2338014+18543−2338014+18543+36−8+2338014+18543+2338014+185432]\left[- \frac{\sqrt{-16 - 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}} - \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + \frac{36}{\sqrt{-8 + \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}}}}}{2} - \frac{\sqrt{-8 + \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}}}{2}, - \frac{\sqrt{-8 + \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}}}{2} + \frac{\sqrt{-16 - 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}} - \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + \frac{36}{\sqrt{-8 + \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}}}}}{2}\right]
Convex at the intervals
(−∞,−−16−2338014+18543−2338014+18543+36−8+2338014+18543+2338014+185432−−8+2338014+18543+2338014+185432]∪[−−8+2338014+18543+2338014+185432+−16−2338014+18543−2338014+18543+36−8+2338014+18543+2338014+185432,∞)\left(-\infty, - \frac{\sqrt{-16 - 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}} - \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + \frac{36}{\sqrt{-8 + \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}}}}}{2} - \frac{\sqrt{-8 + \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}}}{2}\right] \cup \left[- \frac{\sqrt{-8 + \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}}}{2} + \frac{\sqrt{-16 - 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}} - \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + \frac{36}{\sqrt{-8 + \frac{2}{\sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}} + 2 \sqrt[3]{\frac{3 \sqrt{3801}}{4} + \frac{185}{4}}}}}}{2}, \infty\right)
Vertical asymptotes
Have:
x1=0x_{1} = 0
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−3x)+3)=∞\lim_{x \to -\infty} \log{\left(\left(x - \frac{3}{x}\right) + 3 \right)} = \infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞log⁡((x−3x)+3)=∞\lim_{x \to \infty} \log{\left(\left(x - \frac{3}{x}\right) + 3 \right)} = \infty
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of log(x - 3/x + 3), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(log⁡((x−3x)+3)x)=0\lim_{x \to -\infty}\left(\frac{\log{\left(\left(x - \frac{3}{x}\right) + 3 \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(log⁡((x−3x)+3)x)=0\lim_{x \to \infty}\left(\frac{\log{\left(\left(x - \frac{3}{x}\right) + 3 \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:
log⁡((x−3x)+3)=log⁡(−x+3+3x)\log{\left(\left(x - \frac{3}{x}\right) + 3 \right)} = \log{\left(- x + 3 + \frac{3}{x} \right)}
- No
log⁡((x−3x)+3)=−log⁡(−x+3+3x)\log{\left(\left(x - \frac{3}{x}\right) + 3 \right)} = - \log{\left(- x + 3 + \frac{3}{x} \right)}
- No
so, the function
not is
neither even, nor odd
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