Mister Exam
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-
How to use it?
- Graphing y =:
- 2x^3-3x^2+1
- 2x^3+3x^2-1
- 2x^3-3x^2-16
- -2x^2+x+1
-
Identical expressions
- - two (x+ one)/(x^ two +2x)
- minus 2(x plus 1) divide by (x squared plus 2x)
- minus two (x plus one) divide by (x to the power of two plus 2x)
- -2(x+1)/(x2+2x)
- -2x+1/x2+2x
- -2(x+1)/(x²+2x)
- -2(x+1)/(x to the power of 2+2x)
- -2x+1/x^2+2x
- -2(x+1) divide by (x^2+2x)
-
Similar expressions
- 2(x+1)/(x^2+2x)
- -2(x+1)/(x^2-2x)
- -2(x-1)/(x^2+2x)
Graphing y = -2(x+1)/(x^2+2x)
The solution
You have entered
[src]
-2*(x + 1)
f(x) = ----------
2
x + 2*x
$$f{\left(x \right)} = - \frac{2 \left(x + 1\right)}{x^{2} + 2 x}$$
f = -2*(x + 1)/(x^2 + 2*x)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -2$$
$$x_{2} = 0$$
$$x_{1} = -2$$
$$x_{2} = 0$$
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:
$$- \frac{2 \left(x + 1\right)}{x^{2} + 2 x} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = -1$$
Numerical solution
$$x_{1} = -1$$
so we need to solve the equation:
$$- \frac{2 \left(x + 1\right)}{x^{2} + 2 x} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = -1$$
Numerical solution
$$x_{1} = -1$$
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
$$- \frac{2 \left(- 2 x - 2\right) \left(x + 1\right)}{\left(x^{2} + 2 x\right)^{2}} - \frac{2}{x^{2} + 2 x} = 0$$
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
$$- \frac{2 \left(- 2 x - 2\right) \left(x + 1\right)}{\left(x^{2} + 2 x\right)^{2}} - \frac{2}{x^{2} + 2 x} = 0$$
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
$$\frac{4 \cdot \left(3 - \frac{4 \left(x + 1\right)^{2}}{x \left(x + 2\right)}\right) \left(x + 1\right)}{x^{2} \left(x + 2\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = -1$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -2$$
$$x_{2} = 0$$
$$\lim_{x \to -2^-}\left(\frac{4 \cdot \left(3 - \frac{4 \left(x + 1\right)^{2}}{x \left(x + 2\right)}\right) \left(x + 1\right)}{x^{2} \left(x + 2\right)^{2}}\right) = \infty$$
Let's take the limit
$$\lim_{x \to -2^+}\left(\frac{4 \cdot \left(3 - \frac{4 \left(x + 1\right)^{2}}{x \left(x + 2\right)}\right) \left(x + 1\right)}{x^{2} \left(x + 2\right)^{2}}\right) = -\infty$$
Let's take the limit
- the limits are not equal, so
$$x_{1} = -2$$
- is an inflection point
$$\lim_{x \to 0^-}\left(\frac{4 \cdot \left(3 - \frac{4 \left(x + 1\right)^{2}}{x \left(x + 2\right)}\right) \left(x + 1\right)}{x^{2} \left(x + 2\right)^{2}}\right) = \infty$$
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{4 \cdot \left(3 - \frac{4 \left(x + 1\right)^{2}}{x \left(x + 2\right)}\right) \left(x + 1\right)}{x^{2} \left(x + 2\right)^{2}}\right) = -\infty$$
Let's take the limit
- the limits are not equal, so
$$x_{2} = 0$$
- 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
$$\left[-1, \infty\right)$$
Convex at the intervals
$$\left(-\infty, -1\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{4 \cdot \left(3 - \frac{4 \left(x + 1\right)^{2}}{x \left(x + 2\right)}\right) \left(x + 1\right)}{x^{2} \left(x + 2\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = -1$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -2$$
$$x_{2} = 0$$
$$\lim_{x \to -2^-}\left(\frac{4 \cdot \left(3 - \frac{4 \left(x + 1\right)^{2}}{x \left(x + 2\right)}\right) \left(x + 1\right)}{x^{2} \left(x + 2\right)^{2}}\right) = \infty$$
Let's take the limit
$$\lim_{x \to -2^+}\left(\frac{4 \cdot \left(3 - \frac{4 \left(x + 1\right)^{2}}{x \left(x + 2\right)}\right) \left(x + 1\right)}{x^{2} \left(x + 2\right)^{2}}\right) = -\infty$$
Let's take the limit
- the limits are not equal, so
$$x_{1} = -2$$
- is an inflection point
$$\lim_{x \to 0^-}\left(\frac{4 \cdot \left(3 - \frac{4 \left(x + 1\right)^{2}}{x \left(x + 2\right)}\right) \left(x + 1\right)}{x^{2} \left(x + 2\right)^{2}}\right) = \infty$$
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{4 \cdot \left(3 - \frac{4 \left(x + 1\right)^{2}}{x \left(x + 2\right)}\right) \left(x + 1\right)}{x^{2} \left(x + 2\right)^{2}}\right) = -\infty$$
Let's take the limit
- the limits are not equal, so
$$x_{2} = 0$$
- 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
$$\left[-1, \infty\right)$$
Convex at the intervals
$$\left(-\infty, -1\right]$$
Vertical asymptotes
Have:
$$x_{1} = -2$$
$$x_{2} = 0$$
$$x_{1} = -2$$
$$x_{2} = 0$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(- \frac{2 \left(x + 1\right)}{x^{2} + 2 x}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(- \frac{2 \left(x + 1\right)}{x^{2} + 2 x}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty}\left(- \frac{2 \left(x + 1\right)}{x^{2} + 2 x}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(- \frac{2 \left(x + 1\right)}{x^{2} + 2 x}\right) = 0$$
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 + 1)/(x^2 + 2*x), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(- \frac{2 \left(x + 1\right)}{x \left(x^{2} + 2 x\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(- \frac{2 \left(x + 1\right)}{x \left(x^{2} + 2 x\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(- \frac{2 \left(x + 1\right)}{x \left(x^{2} + 2 x\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(- \frac{2 \left(x + 1\right)}{x \left(x^{2} + 2 x\right)}\right) = 0$$
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:
$$- \frac{2 \left(x + 1\right)}{x^{2} + 2 x} = - \frac{2 \cdot \left(- x + 1\right)}{x^{2} - 2 x}$$
- No
$$- \frac{2 \left(x + 1\right)}{x^{2} + 2 x} = \frac{2 \cdot \left(- x + 1\right)}{x^{2} - 2 x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$- \frac{2 \left(x + 1\right)}{x^{2} + 2 x} = - \frac{2 \cdot \left(- x + 1\right)}{x^{2} - 2 x}$$
- No
$$- \frac{2 \left(x + 1\right)}{x^{2} + 2 x} = \frac{2 \cdot \left(- x + 1\right)}{x^{2} - 2 x}$$
- No
so, the function
not is
neither even, nor odd
The graph