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Plotting a function graph/ x+arcctg(x/2)

Graphing y = x+arcctg(x/2)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
               /x\
f(x) = x + acot|-|
               \2/
f(x)=x+acot⁡(x2)f{\left(x \right)} = x + \operatorname{acot}{\left(\frac{x}{2} \right)} 
f = x + acot(x/2)
The graph of the function
02468-8-6-4-2-1010-2020
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:
x+acot⁡(x2)=0x + \operatorname{acot}{\left(\frac{x}{2} \right)} = 0
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to x + acot(x/2).
acot⁡(02)\operatorname{acot}{\left(\frac{0}{2} \right)}
The result:
f(0)=π2f{\left(0 \right)} = \frac{\pi}{2}
The point:
(0, pi/2)
Extrema of the function
In order to find the extrema, we need to solve the equation
ddxf(x)=0\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:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
the first derivative
1−12(x24+1)=01 - \frac{1}{2 \left(\frac{x^{2}}{4} + 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
d2dx2f(x)=0\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:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
x4(x24+1)2=0\frac{x}{4 \left(\frac{x^{2}}{4} + 1\right)^{2}} = 0
Solve this equation
The roots of this equation
x1=0x_{1} = 0

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
[0,∞)\left[0, \infty\right)
Convex at the intervals
(−∞,0]\left(-\infty, 0\right]
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(x+acot⁡(x2))=−∞\lim_{x \to -\infty}\left(x + \operatorname{acot}{\left(\frac{x}{2} \right)}\right) = -\infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞(x+acot⁡(x2))=∞\lim_{x \to \infty}\left(x + \operatorname{acot}{\left(\frac{x}{2} \right)}\right) = \infty
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of x + acot(x/2), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(x+acot⁡(x2)x)=1\lim_{x \to -\infty}\left(\frac{x + \operatorname{acot}{\left(\frac{x}{2} \right)}}{x}\right) = 1
Let's take the limit
so,
inclined asymptote equation on the left:
y=xy = x
lim⁡x→∞(x+acot⁡(x2)x)=1\lim_{x \to \infty}\left(\frac{x + \operatorname{acot}{\left(\frac{x}{2} \right)}}{x}\right) = 1
Let's take the limit
so,
inclined asymptote equation on the right:
y=xy = x
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:
x+acot⁡(x2)=−x−acot⁡(x2)x + \operatorname{acot}{\left(\frac{x}{2} \right)} = - x - \operatorname{acot}{\left(\frac{x}{2} \right)}
- No
x+acot⁡(x2)=x+acot⁡(x2)x + \operatorname{acot}{\left(\frac{x}{2} \right)} = x + \operatorname{acot}{\left(\frac{x}{2} \right)}
- No
so, the function
not is
neither even, nor odd
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