Mister Exam
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-
How to use it?
- Graphing y =:
- ((x-1)/(x+1))^3
- 5x^2-3x-1
- 3/2x^2-x^3
- 3x^2-12x
- Integral of d{x}:
-
(2*x+1)/(x+2)
- Derivative of:
-
(2*x+1)/(x+2)
-
Identical expressions
- (two *x+ one)/(x+ two)
- (2 multiply by x plus 1) divide by (x plus 2)
- (two multiply by x plus one) divide by (x plus two)
- (2x+1)/(x+2)
- 2x+1/x+2
- (2*x+1) divide by (x+2)
-
Similar expressions
- ((2x+1)/(x+2)^2)-2
- (2*x-1)/(x+2)
- (2x+1)/(x+2)
- (2*x+1)/(x-2)
- (x^2+2x+1)/(x+2)
- (x^2+2*x+1)/(x+2)
- (x^2-2*x+1)/(x+2)
Graphing y = (2*x+1)/(x+2)
The solution
You have entered
[src]
2*x + 1
f(x) = -------
x + 2
$$f{\left(x \right)} = \frac{2 x + 1}{x + 2}$$
f = (2*x + 1)/(x + 2)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -2$$
$$x_{1} = -2$$
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:
$$\frac{2 x + 1}{x + 2} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - \frac{1}{2}$$
Numerical solution
$$x_{1} = -0.5$$
so we need to solve the equation:
$$\frac{2 x + 1}{x + 2} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - \frac{1}{2}$$
Numerical solution
$$x_{1} = -0.5$$
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
$$\frac{2}{x + 2} - \frac{2 x + 1}{\left(x + 2\right)^{2}} = 0$$
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
$$\frac{2}{x + 2} - \frac{2 x + 1}{\left(x + 2\right)^{2}} = 0$$
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
$$\frac{2 \left(-2 + \frac{2 x + 1}{x + 2}\right)}{\left(x + 2\right)^{2}} = 0$$
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
$$\frac{2 \left(-2 + \frac{2 x + 1}{x + 2}\right)}{\left(x + 2\right)^{2}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
$$x_{1} = -2$$
$$x_{1} = -2$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{2 x + 1}{x + 2}\right) = 2$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 2$$
$$\lim_{x \to \infty}\left(\frac{2 x + 1}{x + 2}\right) = 2$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 2$$
$$\lim_{x \to -\infty}\left(\frac{2 x + 1}{x + 2}\right) = 2$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 2$$
$$\lim_{x \to \infty}\left(\frac{2 x + 1}{x + 2}\right) = 2$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 2$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (2*x + 1)/(x + 2), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{2 x + 1}{x \left(x + 2\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{2 x + 1}{x \left(x + 2\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{2 x + 1}{x \left(x + 2\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{2 x + 1}{x \left(x + 2\right)}\right) = 0$$
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:
$$\frac{2 x + 1}{x + 2} = \frac{1 - 2 x}{2 - x}$$
- No
$$\frac{2 x + 1}{x + 2} = - \frac{1 - 2 x}{2 - x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{2 x + 1}{x + 2} = \frac{1 - 2 x}{2 - x}$$
- No
$$\frac{2 x + 1}{x + 2} = - \frac{1 - 2 x}{2 - x}$$
- No
so, the function
not is
neither even, nor odd