Mister Exam
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-
How to use it?
- Graphing y =:
- 2x^3-3x^2-12x+1
-
y=|x-2|-|x+1|+x-2
- -1/3x^3+4x
- x^4-4x^2+5
-
Identical expressions
- arccot(x- one)/ two
- arc cotangent of (x minus 1) divide by 2
- arc cotangent of (x minus one) divide by two
- arccotx-1/2
- arccot(x-1) divide by 2
-
Similar expressions
- arccot(x+1)/2
Graphing y = arccot(x-1)/2
The solution
You have entered
[src]
acot(x - 1)
f(x) = -----------
2
$$f{\left(x \right)} = \frac{\operatorname{acot}{\left(x - 1 \right)}}{2}$$
f = acot(x - 1)/2
The graph of the function
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:
$$\frac{\operatorname{acot}{\left(x - 1 \right)}}{2} = 0$$
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:
$$\frac{\operatorname{acot}{\left(x - 1 \right)}}{2} = 0$$
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
$$- \frac{1}{2 \left(\left(x - 1\right)^{2} + 1\right)} = 0$$
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
$$- \frac{1}{2 \left(\left(x - 1\right)^{2} + 1\right)} = 0$$
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
$$\frac{x - 1}{\left(\left(x - 1\right)^{2} + 1\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 1$$
Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
$$\left[1, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 1\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{x - 1}{\left(\left(x - 1\right)^{2} + 1\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 1$$
Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
$$\left[1, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 1\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{\operatorname{acot}{\left(x - 1 \right)}}{2}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{\operatorname{acot}{\left(x - 1 \right)}}{2}\right) = - \frac{\pi}{2}$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = - \frac{\pi}{2}$$
$$\lim_{x \to -\infty}\left(\frac{\operatorname{acot}{\left(x - 1 \right)}}{2}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{\operatorname{acot}{\left(x - 1 \right)}}{2}\right) = - \frac{\pi}{2}$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = - \frac{\pi}{2}$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of acot(x - 1)/2, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\operatorname{acot}{\left(x - 1 \right)}}{2 x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\operatorname{acot}{\left(x - 1 \right)}}{2 x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\operatorname{acot}{\left(x - 1 \right)}}{2 x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\operatorname{acot}{\left(x - 1 \right)}}{2 x}\right) = 0$$
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:
$$\frac{\operatorname{acot}{\left(x - 1 \right)}}{2} = - \frac{\operatorname{acot}{\left(x + 1 \right)}}{2}$$
- No
$$\frac{\operatorname{acot}{\left(x - 1 \right)}}{2} = \frac{\operatorname{acot}{\left(x + 1 \right)}}{2}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{\operatorname{acot}{\left(x - 1 \right)}}{2} = - \frac{\operatorname{acot}{\left(x + 1 \right)}}{2}$$
- No
$$\frac{\operatorname{acot}{\left(x - 1 \right)}}{2} = \frac{\operatorname{acot}{\left(x + 1 \right)}}{2}$$
- No
so, the function
not is
neither even, nor odd