Mister Exam

Other calculators

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

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

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
           /1 + x\
f(x) = -log|-----|
           \1 - x/
f(x)=−log⁡(x+11−x)f{\left(x \right)} = - \log{\left(\frac{x + 1}{1 - x} \right)} 
f = -log((x + 1)/(1 - x))
The graph of the function
-1.0-0.8-0.6-0.4-0.21.00.00.20.40.60.8-2020
The domain of the function
The points at which the function is not precisely defined:
x1=1x_{1} = 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)=0- \log{\left(\frac{x + 1}{1 - x} \right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=0x_{1} = 0
Numerical solution
x1=0x_{1} = 0
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)).
−log⁡(11−0)- \log{\left(\frac{1}{1 - 0} \right)}
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
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+1=0- \frac{\left(1 - x\right) \left(\frac{1}{1 - x} + \frac{x + 1}{\left(1 - x\right)^{2}}\right)}{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
(1−x+1x−1)(1x+1+1x−1)x+1=0\frac{\left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1}{x + 1} + \frac{1}{x - 1}\right)}{x + 1} = 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=1x_{1} = 1

lim⁡x→1−((1−x+1x−1)(1x+1+1x−1)x+1)=−∞\lim_{x \to 1^-}\left(\frac{\left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1}{x + 1} + \frac{1}{x - 1}\right)}{x + 1}\right) = -\infty
lim⁡x→1+((1−x+1x−1)(1x+1+1x−1)x+1)=−∞\lim_{x \to 1^+}\left(\frac{\left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1}{x + 1} + \frac{1}{x - 1}\right)}{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
(−∞,0]\left(-\infty, 0\right]
Convex at the intervals
[0,∞)\left[0, \infty\right)
Vertical asymptotes
Have:
x1=1x_{1} = 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))=−iπ\lim_{x \to -\infty}\left(- \log{\left(\frac{x + 1}{1 - x} \right)}\right) = - i \pi
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=−iπy = - i \pi
lim⁡x→∞(−log⁡(x+11−x))=−iπ\lim_{x \to \infty}\left(- \log{\left(\frac{x + 1}{1 - x} \right)}\right) = - i \pi
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=−iπy = - i \pi
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of -log((1 + x)/(1 - x)), divided by x 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,
inclined coincides with the horizontal asymptote on the right
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,
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)=−log⁡(1−xx+1)- \log{\left(\frac{x + 1}{1 - x} \right)} = - \log{\left(\frac{1 - x}{x + 1} \right)}
- No
−log⁡(x+11−x)=log⁡(1−xx+1)- \log{\left(\frac{x + 1}{1 - x} \right)} = \log{\left(\frac{1 - x}{x + 1} \right)}
- No
so, the function
not is
neither even, nor odd
    © Mister Exam – Calculator