Mister Exam
Other calculators
- Graph of implicit function
- Derivative Step by Step
- Integral Step by Step
- Graph of parametric function
- Graph in Polar Coordinates
- Surface defined parametrically
- Surface defined by equation
- Curve in 3D space defined parametrically
- Area between lines
- The intersection points of functions
-
How to use it?
- Graphing y =:
- x^3+x^2-x+1
- x^2+3x-18
- x²-2x+8
- x^2+2x+1
- Integral of d{x}:
-
x/((x-2)(x+3))
-
Identical expressions
- x/((x- two)(x+ three))
- x divide by ((x minus 2)(x plus 3))
- x divide by ((x minus two)(x plus three))
- x/x-2x+3
- x divide by ((x-2)(x+3))
-
Similar expressions
- x/((x-2)(x-3))
- x/(x-2)(x+3)
- x/((x+2)(x+3))
Graphing y = x/((x-2)(x+3))
The solution
You have entered
[src]
x
f(x) = ---------------
(x - 2)*(x + 3)
$$f{\left(x \right)} = \frac{x}{\left(x - 2\right) \left(x + 3\right)}$$
f = x/(((x - 2)*(x + 3)))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -3$$
$$x_{2} = 2$$
$$x_{1} = -3$$
$$x_{2} = 2$$
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:
$$\frac{x}{\left(x - 2\right) \left(x + 3\right)} = 0$$
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:
$$\frac{x}{\left(x - 2\right) \left(x + 3\right)} = 0$$
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
$$\frac{x \left(- 2 x - 1\right)}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}} + \frac{1}{\left(x - 2\right) \left(x + 3\right)} = 0$$
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
$$\frac{x \left(- 2 x - 1\right)}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}} + \frac{1}{\left(x - 2\right) \left(x + 3\right)} = 0$$
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
$$\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \sqrt[3]{18} + \sqrt[3]{12}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -3$$
$$x_{2} = 2$$
$$\lim_{x \to -3^-}\left(\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}}\right) = -\infty$$
$$\lim_{x \to -3^+}\left(\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = -3$$
- is an inflection point
$$\lim_{x \to 2^-}\left(\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}}\right) = -\infty$$
$$\lim_{x \to 2^+}\left(\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{2} = 2$$
- 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
$$\left(-\infty, - \sqrt[3]{18} + \sqrt[3]{12}\right]$$
Convex at the intervals
$$\left[- \sqrt[3]{18} + \sqrt[3]{12}, \infty\right)$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \sqrt[3]{18} + \sqrt[3]{12}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -3$$
$$x_{2} = 2$$
$$\lim_{x \to -3^-}\left(\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}}\right) = -\infty$$
$$\lim_{x \to -3^+}\left(\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = -3$$
- is an inflection point
$$\lim_{x \to 2^-}\left(\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}}\right) = -\infty$$
$$\lim_{x \to 2^+}\left(\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{2} = 2$$
- 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
$$\left(-\infty, - \sqrt[3]{18} + \sqrt[3]{12}\right]$$
Convex at the intervals
$$\left[- \sqrt[3]{18} + \sqrt[3]{12}, \infty\right)$$
Vertical asymptotes
Have:
$$x_{1} = -3$$
$$x_{2} = 2$$
$$x_{1} = -3$$
$$x_{2} = 2$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{x}{\left(x - 2\right) \left(x + 3\right)}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{x}{\left(x - 2\right) \left(x + 3\right)}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty}\left(\frac{x}{\left(x - 2\right) \left(x + 3\right)}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{x}{\left(x - 2\right) \left(x + 3\right)}\right) = 0$$
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/(((x - 2)*(x + 3))), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{1}{\left(x - 2\right) \left(x + 3\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{1}{\left(x - 2\right) \left(x + 3\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{1}{\left(x - 2\right) \left(x + 3\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{1}{\left(x - 2\right) \left(x + 3\right)}\right) = 0$$
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:
$$\frac{x}{\left(x - 2\right) \left(x + 3\right)} = - \frac{x}{\left(3 - x\right) \left(- x - 2\right)}$$
- No
$$\frac{x}{\left(x - 2\right) \left(x + 3\right)} = \frac{x}{\left(3 - x\right) \left(- x - 2\right)}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{x}{\left(x - 2\right) \left(x + 3\right)} = - \frac{x}{\left(3 - x\right) \left(- x - 2\right)}$$
- No
$$\frac{x}{\left(x - 2\right) \left(x + 3\right)} = \frac{x}{\left(3 - x\right) \left(- x - 2\right)}$$
- No
so, the function
not is
neither even, nor odd