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  • How to use it?

  • Graphing y =:
  • x^3/(2(x+1)^2)
  • x^2*(x-2)^2
  • x/(1+x)^2
  • (x^3+2x^2)/(x-1)^2

  • Identical expressions

  • (x^ two - one)/(x*(x+ one))
  • (x squared minus 1) divide by (x multiply by (x plus 1))
  • (x to the power of two minus one) divide by (x multiply by (x plus one))
  • (x2-1)/(x*(x+1))
  • x2-1/x*x+1
  • (x²-1)/(x*(x+1))
  • (x to the power of 2-1)/(x*(x+1))
  • (x^2-1)/(x(x+1))
  • (x2-1)/(x(x+1))
  • x2-1/xx+1
  • x^2-1/xx+1
  • (x^2-1) divide by (x*(x+1))

  • Similar expressions

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

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

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
          2     
         x  - 1 
f(x) = ---------
       x*(x + 1)
f(x)=x2−1x(x+1)f{\left(x \right)} = \frac{x^{2} - 1}{x \left(x + 1\right)} 
f = (x^2 - 1)/((x*(x + 1)))
The graph of the function
02468-8-6-4-2-1010-5050
The domain of the function
The points at which the function is not precisely defined:
x1=−1x_{1} = -1
x2=0x_{2} = 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:
x2−1x(x+1)=0\frac{x^{2} - 1}{x \left(x + 1\right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=1x_{1} = 1
Numerical solution
x1=1x_{1} = 1
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (x^2 - 1)/((x*(x + 1))).
−1+020\frac{-1 + 0^{2}}{0}
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
2x1x(x+1)+(−2x−1)(x2−1)x2(x+1)2=02 x \frac{1}{x \left(x + 1\right)} + \frac{\left(- 2 x - 1\right) \left(x^{2} - 1\right)}{x^{2} \left(x + 1\right)^{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
2−4(2x+1)x+1+(x2−1)((2x+1)(1x+1+1x)−2+2x+1x+1+2x+1x)x(x+1)x(x+1)=0\frac{2 - \frac{4 \left(2 x + 1\right)}{x + 1} + \frac{\left(x^{2} - 1\right) \left(\left(2 x + 1\right) \left(\frac{1}{x + 1} + \frac{1}{x}\right) - 2 + \frac{2 x + 1}{x + 1} + \frac{2 x + 1}{x}\right)}{x \left(x + 1\right)}}{x \left(x + 1\right)} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
x1=−1x_{1} = -1
x2=0x_{2} = 0
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(x2−1x(x+1))=1\lim_{x \to -\infty}\left(\frac{x^{2} - 1}{x \left(x + 1\right)}\right) = 1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=1y = 1
lim⁡x→∞(x2−1x(x+1))=1\lim_{x \to \infty}\left(\frac{x^{2} - 1}{x \left(x + 1\right)}\right) = 1
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=1y = 1
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (x^2 - 1)/((x*(x + 1))), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(1x(x+1)(x2−1)x)=0\lim_{x \to -\infty}\left(\frac{\frac{1}{x \left(x + 1\right)} \left(x^{2} - 1\right)}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(1x(x+1)(x2−1)x)=0\lim_{x \to \infty}\left(\frac{\frac{1}{x \left(x + 1\right)} \left(x^{2} - 1\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:
x2−1x(x+1)=−x2−1x(1−x)\frac{x^{2} - 1}{x \left(x + 1\right)} = - \frac{x^{2} - 1}{x \left(1 - x\right)}
- No
x2−1x(x+1)=x2−1x(1−x)\frac{x^{2} - 1}{x \left(x + 1\right)} = \frac{x^{2} - 1}{x \left(1 - x\right)}
- No
so, the function
not is
neither even, nor odd
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