Mister Exam
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-
How to use it?
- Graphing y =:
- 1/3x^3-5/2x^2+6x
- x/(1-x^2)
- -2x^2+3x+5
- (x^2-1)/(x+2)
- Integral of d{x}:
-
x/(1-x^2)
- Derivative of:
-
x/(1-x^2)
- Limit of the function:
-
x/(1-x^2)
-
Identical expressions
- x/(one -x^ two)
- x divide by (1 minus x squared )
- x divide by (one minus x to the power of two)
- x/(1-x2)
- x/1-x2
- x/(1-x²)
- x/(1-x to the power of 2)
- x/1-x^2
- x divide by (1-x^2)
-
Similar expressions
- x/(1+x^2)
- (4-2*x)/(1-x^2)
Graphing y = x/(1-x^2)
The solution
You have entered
[src]
x
f(x) = ------
2
1 - x
$$f{\left(x \right)} = \frac{x}{1 - x^{2}}$$
f = x/(1 - x^2)
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:
$$\frac{x}{1 - x^{2}} = 0$$
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:
$$\frac{x}{1 - x^{2}} = 0$$
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
$$\frac{2 x^{2}}{\left(1 - x^{2}\right)^{2}} + \frac{1}{1 - x^{2}} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(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
$$\frac{2 x^{2}}{\left(1 - x^{2}\right)^{2}} + \frac{1}{1 - x^{2}} = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}} = 0$$
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:
$$x_{1} = -1$$
$$x_{2} = 1$$
$$\lim_{x \to -1^-}\left(\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = \infty$$
$$\lim_{x \to -1^+}\left(\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = -\infty$$
- the limits are not equal, so
$$x_{1} = -1$$
- is an inflection point
$$\lim_{x \to 1^-}\left(\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = \infty$$
$$\lim_{x \to 1^+}\left(\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = -\infty$$
- the limits are not equal, so
$$x_{2} = 1$$
- 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
$$\left(-\infty, 0\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}} = 0$$
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:
$$x_{1} = -1$$
$$x_{2} = 1$$
$$\lim_{x \to -1^-}\left(\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = \infty$$
$$\lim_{x \to -1^+}\left(\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = -\infty$$
- the limits are not equal, so
$$x_{1} = -1$$
- is an inflection point
$$\lim_{x \to 1^-}\left(\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = \infty$$
$$\lim_{x \to 1^+}\left(\frac{2 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3\right)}{\left(x^{2} - 1\right)^{2}}\right) = -\infty$$
- the limits are not equal, so
$$x_{2} = 1$$
- 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:
$$x_{1} = -1$$
$$x_{2} = 1$$
$$x_{1} = -1$$
$$x_{2} = 1$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{x}{1 - x^{2}}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{x}{1 - x^{2}}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty}\left(\frac{x}{1 - x^{2}}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{x}{1 - x^{2}}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
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
$$\lim_{x \to -\infty} \frac{1}{1 - x^{2}} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty} \frac{1}{1 - x^{2}} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty} \frac{1}{1 - x^{2}} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty} \frac{1}{1 - x^{2}} = 0$$
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:
$$\frac{x}{1 - x^{2}} = - \frac{x}{1 - x^{2}}$$
- No
$$\frac{x}{1 - x^{2}} = \frac{x}{1 - x^{2}}$$
- Yes
so, the function
is
odd
So, check:
$$\frac{x}{1 - x^{2}} = - \frac{x}{1 - x^{2}}$$
- No
$$\frac{x}{1 - x^{2}} = \frac{x}{1 - x^{2}}$$
- Yes
so, the function
is
odd