Mister Exam
Other calculators
- Graph of implicit function
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- Curve in 3D space defined parametrically
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- The intersection points of functions
-
How to use it?
- Graphing y =:
- x^2-|x-x^2|
- (x^2+x-5)/(x-2)
- -x^2+6x-3
- x/(x-4)
- Integral of d{x}:
-
1/(x+2)
- Derivative of:
-
1/(x+2)
-
Identical expressions
- one /(x+ two)
- 1 divide by (x plus 2)
- one divide by (x plus two)
- 1/x+2
- 1 divide by (x+2)
-
Similar expressions
- tgx/(e^(1/x)+2)
- (2-x)(x+1)/x+2
- (x-1)/(x+2)
- 1/(x-2)
- 1/x+2*x
Graphing y = 1/(x+2)
The solution
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:
$$\frac{1}{x + 2} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
so we need to solve the equation:
$$\frac{1}{x + 2} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
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
$$- \frac{1}{\left(x + 2\right)^{2}} = 0$$
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
$$- \frac{1}{\left(x + 2\right)^{2}} = 0$$
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
$$\frac{2}{\left(x + 2\right)^{3}} = 0$$
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
$$\frac{2}{\left(x + 2\right)^{3}} = 0$$
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
$$\lim_{x \to -\infty} \frac{1}{x + 2} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty} \frac{1}{x + 2} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty} \frac{1}{x + 2} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty} \frac{1}{x + 2} = 0$$
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 1/(x + 2), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{1}{x \left(x + 2\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{1}{x \left(x + 2\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{1}{x \left(x + 2\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{1}{x \left(x + 2\right)}\right) = 0$$
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:
$$\frac{1}{x + 2} = \frac{1}{2 - x}$$
- No
$$\frac{1}{x + 2} = - \frac{1}{2 - x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{1}{x + 2} = \frac{1}{2 - x}$$
- No
$$\frac{1}{x + 2} = - \frac{1}{2 - x}$$
- No
so, the function
not is
neither even, nor odd