Mister Exam
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-
How to use it?
- Graphing y =:
- x^2-6x+9
- -x^2+6x-3
- x^2-2x+8
- (x+5)^2
- Derivative of:
-
2*x/(x^2-1)
- How do you in partial fractions?:
- 2*x/(x^2-1)
- Integral of d{x}:
-
2*x/(x^2-1)
-
Identical expressions
- two *x/(x^ two - one)
- 2 multiply by x divide by (x squared minus 1)
- two multiply by x divide by (x to the power of two minus one)
- 2*x/(x2-1)
- 2*x/x2-1
- 2*x/(x²-1)
- 2*x/(x to the power of 2-1)
- 2x/(x^2-1)
- 2x/(x2-1)
- 2x/x2-1
- 2x/x^2-1
- 2*x divide by (x^2-1)
-
Similar expressions
- 2*x/(x^2+1)
- -2x/(x^2-1)^2
Graphing y = 2*x/(x^2-1)
The solution
You have entered
[src]
2*x
f(x) = ------
2
x - 1
$$f{\left(x \right)} = \frac{2 x}{x^{2} - 1}$$
f = (2*x)/(x^2 - 1)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -1$$
$$x_{2} = 1$$
$$x_{1} = -1$$
$$x_{2} = 1$$
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:
$$\frac{2 x}{x^{2} - 1} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 0$$
Numerical solution
$$x_{1} = 0$$
so we need to solve the equation:
$$\frac{2 x}{x^{2} - 1} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 0$$
Numerical solution
$$x_{1} = 0$$
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
$$- \frac{4 x^{2}}{\left(x^{2} - 1\right)^{2}} + \frac{2}{x^{2} - 1} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(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
$$- \frac{4 x^{2}}{\left(x^{2} - 1\right)^{2}} + \frac{2}{x^{2} - 1} = 0$$
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
$$\frac{4 x \left(\frac{4 x^{2}}{x^{2} - 1} - 3\right)}{\left(x^{2} - 1\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -1$$
$$x_{2} = 1$$
$$\lim_{x \to -1^-}\left(\frac{4 x \left(\frac{4 x^{2}}{x^{2} - 1} - 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = -\infty$$
$$\lim_{x \to -1^+}\left(\frac{4 x \left(\frac{4 x^{2}}{x^{2} - 1} - 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = -1$$
- is an inflection point
$$\lim_{x \to 1^-}\left(\frac{4 x \left(\frac{4 x^{2}}{x^{2} - 1} - 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = -\infty$$
$$\lim_{x \to 1^+}\left(\frac{4 x \left(\frac{4 x^{2}}{x^{2} - 1} - 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{2} = 1$$
- 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
$$\left(-\infty, 0\right]$$
Convex at the intervals
$$\left[0, \infty\right)$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{4 x \left(\frac{4 x^{2}}{x^{2} - 1} - 3\right)}{\left(x^{2} - 1\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -1$$
$$x_{2} = 1$$
$$\lim_{x \to -1^-}\left(\frac{4 x \left(\frac{4 x^{2}}{x^{2} - 1} - 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = -\infty$$
$$\lim_{x \to -1^+}\left(\frac{4 x \left(\frac{4 x^{2}}{x^{2} - 1} - 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = -1$$
- is an inflection point
$$\lim_{x \to 1^-}\left(\frac{4 x \left(\frac{4 x^{2}}{x^{2} - 1} - 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = -\infty$$
$$\lim_{x \to 1^+}\left(\frac{4 x \left(\frac{4 x^{2}}{x^{2} - 1} - 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{2} = 1$$
- 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
$$\left(-\infty, 0\right]$$
Convex at the intervals
$$\left[0, \infty\right)$$
Vertical asymptotes
Have:
$$x_{1} = -1$$
$$x_{2} = 1$$
$$x_{1} = -1$$
$$x_{2} = 1$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{2 x}{x^{2} - 1}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{2 x}{x^{2} - 1}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty}\left(\frac{2 x}{x^{2} - 1}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{2 x}{x^{2} - 1}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (2*x)/(x^2 - 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{2}{x^{2} - 1}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{2}{x^{2} - 1}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{2}{x^{2} - 1}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{2}{x^{2} - 1}\right) = 0$$
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:
$$\frac{2 x}{x^{2} - 1} = - \frac{2 x}{x^{2} - 1}$$
- No
$$\frac{2 x}{x^{2} - 1} = \frac{2 x}{x^{2} - 1}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{2 x}{x^{2} - 1} = - \frac{2 x}{x^{2} - 1}$$
- No
$$\frac{2 x}{x^{2} - 1} = \frac{2 x}{x^{2} - 1}$$
- No
so, the function
not is
neither even, nor odd