Mister Exam
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-
How to use it?
- Graphing y =:
- x/(2x+1)
- x^3-9/2x^2+6x-2
-
x/(1-x)
- x-sin(x)
- Integral of d{x}:
-
x/(2x+1)
- Derivative of:
- x/(2x+1)
-
Identical expressions
- x/(2x+ one)
- x divide by (2x plus 1)
- x divide by (2x plus one)
- x/2x+1
- x divide by (2x+1)
-
Similar expressions
- 2^x/(2^(x+1)-1)
- e^x/(2*x+1)
- (8x^(3)+8x)/(2x+1)^3
- x/(2x-1)
- (1-(-1)^x)/(2x+1)
- 3(1-4x)/(2x+1)^2
Graphing y = x/(2x+1)
The solution
You have entered
[src]
x
f(x) = -------
2*x + 1
$$f{\left(x \right)} = \frac{x}{2 x + 1}$$
f = x/(2*x + 1)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -0.5$$
$$x_{1} = -0.5$$
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:
$$\frac{x}{2 x + 1} = 0$$
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:
$$\frac{x}{2 x + 1} = 0$$
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
$$- \frac{2 x}{\left(2 x + 1\right)^{2}} + \frac{1}{2 x + 1} = 0$$
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
$$- \frac{2 x}{\left(2 x + 1\right)^{2}} + \frac{1}{2 x + 1} = 0$$
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
$$\frac{4 \left(\frac{2 x}{2 x + 1} - 1\right)}{\left(2 x + 1\right)^{2}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{4 \left(\frac{2 x}{2 x + 1} - 1\right)}{\left(2 x + 1\right)^{2}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
$$x_{1} = -0.5$$
$$x_{1} = -0.5$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{x}{2 x + 1}\right) = \frac{1}{2}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \frac{1}{2}$$
$$\lim_{x \to \infty}\left(\frac{x}{2 x + 1}\right) = \frac{1}{2}$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = \frac{1}{2}$$
$$\lim_{x \to -\infty}\left(\frac{x}{2 x + 1}\right) = \frac{1}{2}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \frac{1}{2}$$
$$\lim_{x \to \infty}\left(\frac{x}{2 x + 1}\right) = \frac{1}{2}$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = \frac{1}{2}$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of x/(2*x + 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty} \frac{1}{2 x + 1} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty} \frac{1}{2 x + 1} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty} \frac{1}{2 x + 1} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty} \frac{1}{2 x + 1} = 0$$
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:
$$\frac{x}{2 x + 1} = - \frac{x}{1 - 2 x}$$
- No
$$\frac{x}{2 x + 1} = \frac{x}{1 - 2 x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{x}{2 x + 1} = - \frac{x}{1 - 2 x}$$
- No
$$\frac{x}{2 x + 1} = \frac{x}{1 - 2 x}$$
- No
so, the function
not is
neither even, nor odd