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

Graphing y = (1\x)ln((1+x)/(1-x))

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
          /1 + x\
       log|-----|
          \1 - x/
f(x) = ----------
           x
f(x)=log⁡(x+11−x)xf{\left(x \right)} = \frac{\log{\left(\frac{x + 1}{1 - x} \right)}}{x} 
f = log((x + 1)/(1 - x))/x
The graph of the function
02468-8-6-4-2-101004
The domain of the function
The points at which the function is not precisely defined:
x1=0x_{1} = 0
x2=1x_{2} = 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:
log⁡(x+11−x)x=0\frac{\log{\left(\frac{x + 1}{1 - x} \right)}}{x} = 0
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to log((1 + x)/(1 - x))/x.
log⁡(11−0)0\frac{\log{\left(\frac{1}{1 - 0} \right)}}{0}
The result:
f(0)=NaNf{\left(0 \right)} = \text{NaN}
- the solutions of the equation d'not exist
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−x)(11−x+x+1(1−x)2)x(x+1)−log⁡(x+11−x)x2=0\frac{\left(1 - x\right) \left(\frac{1}{1 - x} + \frac{x + 1}{\left(1 - x\right)^{2}}\right)}{x \left(x + 1\right)} - \frac{\log{\left(\frac{x + 1}{1 - x} \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+1x−1)(1x+1+1x−1)x+1−2(1−x+1x−1)x(x+1)+2log⁡(−x+1x−1)x2x=0\frac{- \frac{\left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1}{x + 1} + \frac{1}{x - 1}\right)}{x + 1} - \frac{2 \left(1 - \frac{x + 1}{x - 1}\right)}{x \left(x + 1\right)} + \frac{2 \log{\left(- \frac{x + 1}{x - 1} \right)}}{x^{2}}}{x} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
x1=0x_{1} = 0
x2=1x_{2} = 1
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+11−x)x)=0\lim_{x \to -\infty}\left(\frac{\log{\left(\frac{x + 1}{1 - x} \right)}}{x}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(log⁡(x+11−x)x)=0\lim_{x \to \infty}\left(\frac{\log{\left(\frac{x + 1}{1 - x} \right)}}{x}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=0y = 0
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of log((1 + x)/(1 - x))/x, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(log⁡(x+11−x)x2)=0\lim_{x \to -\infty}\left(\frac{\log{\left(\frac{x + 1}{1 - x} \right)}}{x^{2}}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(log⁡(x+11−x)x2)=0\lim_{x \to \infty}\left(\frac{\log{\left(\frac{x + 1}{1 - x} \right)}}{x^{2}}\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+11−x)x=−log⁡(1−xx+1)x\frac{\log{\left(\frac{x + 1}{1 - x} \right)}}{x} = - \frac{\log{\left(\frac{1 - x}{x + 1} \right)}}{x}
- No
log⁡(x+11−x)x=log⁡(1−xx+1)x\frac{\log{\left(\frac{x + 1}{1 - x} \right)}}{x} = \frac{\log{\left(\frac{1 - x}{x + 1} \right)}}{x}
- No
so, the function
not is
neither even, nor odd
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