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

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

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
         x   
f(x) = ------
            2
       1 - x
f(x)=x1−x2f{\left(x \right)} = \frac{x}{1 - x^{2}} 
f = x/(1 - x^2)
The graph of the function
02468-8-6-4-2-1010-2525
The domain of the function
The points at which the function is not precisely defined:
x1=−1x_{1} = -1
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:
x1−x2=0\frac{x}{1 - x^{2}} = 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 x/(1 - x^2).
01−02\frac{0}{1 - 0^{2}}
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
2x2(1−x2)2+11−x2=0\frac{2 x^{2}}{\left(1 - x^{2}\right)^{2}} + \frac{1}{1 - 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
2x(−4x2x2−1+3)(x2−1)2=0\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}} = 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
x2=1x_{2} = 1

lim⁡x→−1−(2x(−4x2x2−1+3)(x2−1)2)=∞\lim_{x \to -1^-}\left(\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = \infty
lim⁡x→−1+(2x(−4x2x2−1+3)(x2−1)2)=−∞\lim_{x \to -1^+}\left(\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = -\infty
- the limits are not equal, so
x1=−1x_{1} = -1
- is an inflection point
lim⁡x→1−(2x(−4x2x2−1+3)(x2−1)2)=∞\lim_{x \to 1^-}\left(\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = \infty
lim⁡x→1+(2x(−4x2x2−1+3)(x2−1)2)=−∞\lim_{x \to 1^+}\left(\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = -\infty
- the limits are not equal, so
x2=1x_{2} = 1
- is an inflection 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=−1x_{1} = -1
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→−∞(x1−x2)=0\lim_{x \to -\infty}\left(\frac{x}{1 - x^{2}}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(x1−x2)=0\lim_{x \to \infty}\left(\frac{x}{1 - x^{2}}\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 x/(1 - x^2), divided by x at x->+oo and x ->-oo
lim⁡x→−∞11−x2=0\lim_{x \to -\infty} \frac{1}{1 - x^{2}} = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞11−x2=0\lim_{x \to \infty} \frac{1}{1 - x^{2}} = 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:
x1−x2=−x1−x2\frac{x}{1 - x^{2}} = - \frac{x}{1 - x^{2}}
- No
x1−x2=x1−x2\frac{x}{1 - x^{2}} = \frac{x}{1 - x^{2}}
- Yes
so, the function
is
odd
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