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Plotting a function graph/ ln((2x-1)/(x^2-9))

Graphing y = ln((2x-1)/(x^2-9))

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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          /2*x - 1\
f(x) = log|-------|
          |  2    |
          \ x  - 9/
f(x)=log⁡(2x−1x2−9)f{\left(x \right)} = \log{\left(\frac{2 x - 1}{x^{2} - 9} \right)} 
f = log((2*x - 1)/(x^2 - 9))
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=−3x_{1} = -3
x2=3x_{2} = 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:
log⁡(2x−1x2−9)=0\log{\left(\frac{2 x - 1}{x^{2} - 9} \right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−2x_{1} = -2
x2=4x_{2} = 4
Numerical solution
x1=−2x_{1} = -2
x2=4x_{2} = 4
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to log((2*x - 1)/(x^2 - 9)).
log⁡(−1+0⋅2−9+02)\log{\left(\frac{-1 + 0 \cdot 2}{-9 + 0^{2}} \right)}
The result:
f(0)=−log⁡(9)f{\left(0 \right)} = - \log{\left(9 \right)}
The point:
(0, -log(9))
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
(x2−9)(−2x(2x−1)(x2−9)2+2x2−9)2x−1=0\frac{\left(x^{2} - 9\right) \left(- \frac{2 x \left(2 x - 1\right)}{\left(x^{2} - 9\right)^{2}} + \frac{2}{x^{2} - 9}\right)}{2 x - 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(−2x(x(2x−1)x2−9−1)x2−9−−4x2(2x−1)x2−9+6x−1x2−9+2(x(2x−1)x2−9−1)2x−1)2x−1=0\frac{2 \left(- \frac{2 x \left(\frac{x \left(2 x - 1\right)}{x^{2} - 9} - 1\right)}{x^{2} - 9} - \frac{- \frac{4 x^{2} \left(2 x - 1\right)}{x^{2} - 9} + 6 x - 1}{x^{2} - 9} + \frac{2 \left(\frac{x \left(2 x - 1\right)}{x^{2} - 9} - 1\right)}{2 x - 1}\right)}{2 x - 1} = 0
Solve this equation
The roots of this equation
x1=−−703+1225721225270696+114782517283+21225270696+1147825172832+12+−1403−21225270696+114782517283−1225721225270696+114782517283+35−703+1225721225270696+114782517283+21225270696+1147825172832x_{1} = - \frac{\sqrt{- \frac{70}{3} + \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}}}{2} + \frac{1}{2} + \frac{\sqrt{- \frac{140}{3} - 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}} - \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + \frac{35}{\sqrt{- \frac{70}{3} + \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}}}}}{2}
x2=−−1403−21225270696+114782517283−1225721225270696+114782517283+35−703+1225721225270696+114782517283+21225270696+1147825172832−−703+1225721225270696+114782517283+21225270696+1147825172832+12x_{2} = - \frac{\sqrt{- \frac{140}{3} - 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}} - \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + \frac{35}{\sqrt{- \frac{70}{3} + \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}}}}}{2} - \frac{\sqrt{- \frac{70}{3} + \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}}}{2} + \frac{1}{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=−3x_{1} = -3
x2=3x_{2} = 3

lim⁡x→−3−(2(−2x(x(2x−1)x2−9−1)x2−9−−4x2(2x−1)x2−9+6x−1x2−9+2(x(2x−1)x2−9−1)2x−1)2x−1)=∞\lim_{x \to -3^-}\left(\frac{2 \left(- \frac{2 x \left(\frac{x \left(2 x - 1\right)}{x^{2} - 9} - 1\right)}{x^{2} - 9} - \frac{- \frac{4 x^{2} \left(2 x - 1\right)}{x^{2} - 9} + 6 x - 1}{x^{2} - 9} + \frac{2 \left(\frac{x \left(2 x - 1\right)}{x^{2} - 9} - 1\right)}{2 x - 1}\right)}{2 x - 1}\right) = \infty
lim⁡x→−3+(2(−2x(x(2x−1)x2−9−1)x2−9−−4x2(2x−1)x2−9+6x−1x2−9+2(x(2x−1)x2−9−1)2x−1)2x−1)=∞\lim_{x \to -3^+}\left(\frac{2 \left(- \frac{2 x \left(\frac{x \left(2 x - 1\right)}{x^{2} - 9} - 1\right)}{x^{2} - 9} - \frac{- \frac{4 x^{2} \left(2 x - 1\right)}{x^{2} - 9} + 6 x - 1}{x^{2} - 9} + \frac{2 \left(\frac{x \left(2 x - 1\right)}{x^{2} - 9} - 1\right)}{2 x - 1}\right)}{2 x - 1}\right) = \infty
- limits are equal, then skip the corresponding point
lim⁡x→3−(2(−2x(x(2x−1)x2−9−1)x2−9−−4x2(2x−1)x2−9+6x−1x2−9+2(x(2x−1)x2−9−1)2x−1)2x−1)=∞\lim_{x \to 3^-}\left(\frac{2 \left(- \frac{2 x \left(\frac{x \left(2 x - 1\right)}{x^{2} - 9} - 1\right)}{x^{2} - 9} - \frac{- \frac{4 x^{2} \left(2 x - 1\right)}{x^{2} - 9} + 6 x - 1}{x^{2} - 9} + \frac{2 \left(\frac{x \left(2 x - 1\right)}{x^{2} - 9} - 1\right)}{2 x - 1}\right)}{2 x - 1}\right) = \infty
lim⁡x→3+(2(−2x(x(2x−1)x2−9−1)x2−9−−4x2(2x−1)x2−9+6x−1x2−9+2(x(2x−1)x2−9−1)2x−1)2x−1)=∞\lim_{x \to 3^+}\left(\frac{2 \left(- \frac{2 x \left(\frac{x \left(2 x - 1\right)}{x^{2} - 9} - 1\right)}{x^{2} - 9} - \frac{- \frac{4 x^{2} \left(2 x - 1\right)}{x^{2} - 9} + 6 x - 1}{x^{2} - 9} + \frac{2 \left(\frac{x \left(2 x - 1\right)}{x^{2} - 9} - 1\right)}{2 x - 1}\right)}{2 x - 1}\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
(−∞,−−1403−21225270696+114782517283−1225721225270696+114782517283+35−703+1225721225270696+114782517283+21225270696+1147825172832−−703+1225721225270696+114782517283+21225270696+1147825172832+12]∪[−−703+1225721225270696+114782517283+21225270696+1147825172832+12+−1403−21225270696+114782517283−1225721225270696+114782517283+35−703+1225721225270696+114782517283+21225270696+1147825172832,∞)\left(-\infty, - \frac{\sqrt{- \frac{140}{3} - 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}} - \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + \frac{35}{\sqrt{- \frac{70}{3} + \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}}}}}{2} - \frac{\sqrt{- \frac{70}{3} + \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}}}{2} + \frac{1}{2}\right] \cup \left[- \frac{\sqrt{- \frac{70}{3} + \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}}}{2} + \frac{1}{2} + \frac{\sqrt{- \frac{140}{3} - 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}} - \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + \frac{35}{\sqrt{- \frac{70}{3} + \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}}}}}{2}, \infty\right)
Convex at the intervals
[−−1403−21225270696+114782517283−1225721225270696+114782517283+35−703+1225721225270696+114782517283+21225270696+1147825172832−−703+1225721225270696+114782517283+21225270696+1147825172832+12,−−703+1225721225270696+114782517283+21225270696+1147825172832+12+−1403−21225270696+114782517283−1225721225270696+114782517283+35−703+1225721225270696+114782517283+21225270696+1147825172832]\left[- \frac{\sqrt{- \frac{140}{3} - 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}} - \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + \frac{35}{\sqrt{- \frac{70}{3} + \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}}}}}{2} - \frac{\sqrt{- \frac{70}{3} + \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}}}{2} + \frac{1}{2}, - \frac{\sqrt{- \frac{70}{3} + \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}}}{2} + \frac{1}{2} + \frac{\sqrt{- \frac{140}{3} - 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}} - \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + \frac{35}{\sqrt{- \frac{70}{3} + \frac{1225}{72 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}} + 2 \sqrt[3]{\frac{1225 \sqrt{2706}}{96} + \frac{1147825}{1728}}}}}}{2}\right]
Vertical asymptotes
Have:
x1=−3x_{1} = -3
x2=3x_{2} = 3
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞log⁡(2x−1x2−9)=−∞\lim_{x \to -\infty} \log{\left(\frac{2 x - 1}{x^{2} - 9} \right)} = -\infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞log⁡(2x−1x2−9)=−∞\lim_{x \to \infty} \log{\left(\frac{2 x - 1}{x^{2} - 9} \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((2*x - 1)/(x^2 - 9)), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(log⁡(2x−1x2−9)x)=0\lim_{x \to -\infty}\left(\frac{\log{\left(\frac{2 x - 1}{x^{2} - 9} \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(log⁡(2x−1x2−9)x)=0\lim_{x \to \infty}\left(\frac{\log{\left(\frac{2 x - 1}{x^{2} - 9} \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⁡(2x−1x2−9)=log⁡(−2x−1x2−9)\log{\left(\frac{2 x - 1}{x^{2} - 9} \right)} = \log{\left(\frac{- 2 x - 1}{x^{2} - 9} \right)}
- No
log⁡(2x−1x2−9)=−log⁡(−2x−1x2−9)\log{\left(\frac{2 x - 1}{x^{2} - 9} \right)} = - \log{\left(\frac{- 2 x - 1}{x^{2} - 9} \right)}
- No
so, the function
not is
neither even, nor odd
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