Mister Exam
Other calculators
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- 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^2-6x+5
- 4x/(x+1)^2
- x^2-x
-
x/(x-1)
- Sum of series:
-
1/n
- Limit of the function:
-
1/n
-
Identical expressions
- one /n
- 1 divide by n
- one divide by n
-
Similar expressions
- arcsin((n+1)/(n^2+3))
- 1/(n*log(n))
- (n+1)/n
- (((n+2)/(n+1))^(n+1))^(n+1)/(((n+1)/n)^n)^n
- ((-1)^n+1)/n
Graphing y = 1/n
The solution
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$n_{1} = 0$$
$$n_{1} = 0$$
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis N at f = 0
so we need to solve the equation:
$$\frac{1}{n} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis N
so we need to solve the equation:
$$\frac{1}{n} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis N
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d n} f{\left(n \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d n} f{\left(n \right)} = $$
the first derivative
$$- \frac{1}{n^{2}} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
$$\frac{d}{d n} f{\left(n \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d n} f{\left(n \right)} = $$
the first derivative
$$- \frac{1}{n^{2}} = 0$$
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
$$\frac{d^{2}}{d n^{2}} f{\left(n \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d n^{2}} f{\left(n \right)} = $$
the second derivative
$$\frac{2}{n^{3}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
$$\frac{d^{2}}{d n^{2}} f{\left(n \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d n^{2}} f{\left(n \right)} = $$
the second derivative
$$\frac{2}{n^{3}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
$$n_{1} = 0$$
$$n_{1} = 0$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at n->+oo and n->-oo
$$\lim_{n \to -\infty} \frac{1}{n} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{n \to \infty} \frac{1}{n} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{n \to -\infty} \frac{1}{n} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{n \to \infty} \frac{1}{n} = 0$$
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/n, divided by n at n->+oo and n ->-oo
$$\lim_{n \to -\infty} \frac{1}{n^{2}} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{n \to \infty} \frac{1}{n^{2}} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{n \to -\infty} \frac{1}{n^{2}} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{n \to \infty} \frac{1}{n^{2}} = 0$$
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(-n) и f = -f(-n).
So, check:
$$\frac{1}{n} = - \frac{1}{n}$$
- No
$$\frac{1}{n} = \frac{1}{n}$$
- Yes
so, the function
is
odd
So, check:
$$\frac{1}{n} = - \frac{1}{n}$$
- No
$$\frac{1}{n} = \frac{1}{n}$$
- Yes
so, the function
is
odd