Mister Exam

Other calculators

  • How to use it?

  • Graphing y =:
  • (x^5/5)-(x^3/3)
  • ||x-4|-2|
  • |x-3|-|x+3|
  • x^3+x^2-8x+1
  • Derivative of:
  • (x^3-3*x)/(x^2-1) (x^3-3*x)/(x^2-1)
  • Identical expressions

  • (x^ three - three *x)/(x^ two - one)
  • (x cubed minus 3 multiply by x) divide by (x squared minus 1)
  • (x to the power of three minus three multiply by x) divide by (x to the power of two minus one)
  • (x3-3*x)/(x2-1)
  • x3-3*x/x2-1
  • (x³-3*x)/(x²-1)
  • (x to the power of 3-3*x)/(x to the power of 2-1)
  • (x^3-3x)/(x^2-1)
  • (x3-3x)/(x2-1)
  • x3-3x/x2-1
  • x^3-3x/x^2-1
  • (x^3-3*x) divide by (x^2-1)
  • Similar expressions

  • (x^3+3*x)/(x^2-1)
  • (x^3-3*x)/(x^2+1)

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

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src]
        3      
       x  - 3*x
f(x) = --------
         2     
        x  - 1 
$$f{\left(x \right)} = \frac{x^{3} - 3 x}{x^{2} - 1}$$
f = (x^3 - 3*x)/(x^2 - 1)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -1$$
$$x_{2} = 1$$
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:
$$\frac{x^{3} - 3 x}{x^{2} - 1} = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = 0$$
$$x_{2} = - \sqrt{3}$$
$$x_{3} = \sqrt{3}$$
Numerical solution
$$x_{1} = -1.73205080756888$$
$$x_{2} = 1.73205080756888$$
$$x_{3} = 0$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (x^3 - 3*x)/(x^2 - 1).
$$\frac{0^{3} - 0}{-1 + 0^{2}}$$
The result:
$$f{\left(0 \right)} = 0$$
The point:
(0, 0)
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$- \frac{2 x \left(x^{3} - 3 x\right)}{\left(x^{2} - 1\right)^{2}} + \frac{3 x^{2} - 3}{x^{2} - 1} = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$\frac{2 x \left(\frac{\left(x^{2} - 3\right) \left(\frac{4 x^{2}}{x^{2} - 1} - 1\right)}{x^{2} - 1} - 3\right)}{x^{2} - 1} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -1$$
$$x_{2} = 1$$

$$\lim_{x \to -1^-}\left(\frac{2 x \left(\frac{\left(x^{2} - 3\right) \left(\frac{4 x^{2}}{x^{2} - 1} - 1\right)}{x^{2} - 1} - 3\right)}{x^{2} - 1}\right) = \infty$$
$$\lim_{x \to -1^+}\left(\frac{2 x \left(\frac{\left(x^{2} - 3\right) \left(\frac{4 x^{2}}{x^{2} - 1} - 1\right)}{x^{2} - 1} - 3\right)}{x^{2} - 1}\right) = -\infty$$
- the limits are not equal, so
$$x_{1} = -1$$
- is an inflection point
$$\lim_{x \to 1^-}\left(\frac{2 x \left(\frac{\left(x^{2} - 3\right) \left(\frac{4 x^{2}}{x^{2} - 1} - 1\right)}{x^{2} - 1} - 3\right)}{x^{2} - 1}\right) = \infty$$
$$\lim_{x \to 1^+}\left(\frac{2 x \left(\frac{\left(x^{2} - 3\right) \left(\frac{4 x^{2}}{x^{2} - 1} - 1\right)}{x^{2} - 1} - 3\right)}{x^{2} - 1}\right) = -\infty$$
- the limits are not equal, so
$$x_{2} = 1$$
- 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
$$\left[0, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 0\right]$$
Vertical asymptotes
Have:
$$x_{1} = -1$$
$$x_{2} = 1$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{x^{3} - 3 x}{x^{2} - 1}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{x^{3} - 3 x}{x^{2} - 1}\right) = \infty$$
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 (x^3 - 3*x)/(x^2 - 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{x^{3} - 3 x}{x \left(x^{2} - 1\right)}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x$$
$$\lim_{x \to \infty}\left(\frac{x^{3} - 3 x}{x \left(x^{2} - 1\right)}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x$$
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:
$$\frac{x^{3} - 3 x}{x^{2} - 1} = \frac{- x^{3} + 3 x}{x^{2} - 1}$$
- No
$$\frac{x^{3} - 3 x}{x^{2} - 1} = - \frac{- x^{3} + 3 x}{x^{2} - 1}$$
- No
so, the function
not is
neither even, nor odd