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^4/4-x^3/3-x^2
- x^3+6*x^2+9*x
- (x^3-5x^2-6x)/x
- x^3-4x^2
- Integral of d{x}:
-
x/(x+1)
- Derivative of:
-
x/(x+1)
-
Identical expressions
- x/(x+ one)
- x divide by (x plus 1)
- x divide by (x plus one)
- x/x+1
- x divide by (x+1)
-
Similar expressions
- x/(x-1)
- exp(x)/(x+1)
- (x^2-8*x)/(x+1)
- sqrt(x)/(x+100)
- (-cos(x)+(2*sin(x)/(x+1))+(2*cos(x)/(x+1)^2))/(x+1)
- cos(x)/(x+1)
Graphing y = x/(x+1)
The solution
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -1$$
$$x_{1} = -1$$
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:
$$\frac{x}{x + 1} = 0$$
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:
$$\frac{x}{x + 1} = 0$$
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
$$- \frac{x}{\left(x + 1\right)^{2}} + \frac{1}{x + 1} = 0$$
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
$$- \frac{x}{\left(x + 1\right)^{2}} + \frac{1}{x + 1} = 0$$
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
$$\frac{2 \left(\frac{x}{x + 1} - 1\right)}{\left(x + 1\right)^{2}} = 0$$
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
$$\frac{2 \left(\frac{x}{x + 1} - 1\right)}{\left(x + 1\right)^{2}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
$$x_{1} = -1$$
$$x_{1} = -1$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{x}{x + 1}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{x}{x + 1}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
$$\lim_{x \to -\infty}\left(\frac{x}{x + 1}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{x}{x + 1}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of x/(x + 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty} \frac{1}{x + 1} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty} \frac{1}{x + 1} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty} \frac{1}{x + 1} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty} \frac{1}{x + 1} = 0$$
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:
$$\frac{x}{x + 1} = - \frac{x}{1 - x}$$
- No
$$\frac{x}{x + 1} = \frac{x}{1 - x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{x}{x + 1} = - \frac{x}{1 - x}$$
- No
$$\frac{x}{x + 1} = \frac{x}{1 - x}$$
- No
so, the function
not is
neither even, nor odd