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Plotting a function graph/ 2-arccot(x+3)

Graphing y = 2-arccot(x+3)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = 2 - acot(x + 3)
f(x)=2−acot⁡(x+3)f{\left(x \right)} = 2 - \operatorname{acot}{\left(x + 3 \right)} 
f = 2 - acot(x + 3)
The graph of the function
02468-8-6-4-2-101005
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:
2−acot⁡(x+3)=02 - \operatorname{acot}{\left(x + 3 \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 2 - acot(x + 3).
2−acot⁡(3)2 - \operatorname{acot}{\left(3 \right)}
The result:
f(0)=2−acot⁡(3)f{\left(0 \right)} = 2 - \operatorname{acot}{\left(3 \right)}
The point:
(0, 2 - acot(3))
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(x+3)2+1=0\frac{1}{\left(x + 3\right)^{2} + 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
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
−2(x+3)((x+3)2+1)2=0- \frac{2 \left(x + 3\right)}{\left(\left(x + 3\right)^{2} + 1\right)^{2}} = 0
Solve this equation
The roots of this equation
x1=−3x_{1} = -3

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
(−∞,−3]\left(-\infty, -3\right]
Convex at the intervals
[−3,∞)\left[-3, \infty\right)
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(2−acot⁡(x+3))=2−π\lim_{x \to -\infty}\left(2 - \operatorname{acot}{\left(x + 3 \right)}\right) = 2 - \pi
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=2−πy = 2 - \pi
lim⁡x→∞(2−acot⁡(x+3))=2\lim_{x \to \infty}\left(2 - \operatorname{acot}{\left(x + 3 \right)}\right) = 2
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=2y = 2
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 2 - acot(x + 3), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(2−acot⁡(x+3)x)=0\lim_{x \to -\infty}\left(\frac{2 - \operatorname{acot}{\left(x + 3 \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(2−acot⁡(x+3)x)=0\lim_{x \to \infty}\left(\frac{2 - \operatorname{acot}{\left(x + 3 \right)}}{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:
2−acot⁡(x+3)=acot⁡(x−3)+22 - \operatorname{acot}{\left(x + 3 \right)} = \operatorname{acot}{\left(x - 3 \right)} + 2
- No
2−acot⁡(x+3)=−acot⁡(x−3)−22 - \operatorname{acot}{\left(x + 3 \right)} = - \operatorname{acot}{\left(x - 3 \right)} - 2
- No
so, the function
not is
neither even, nor odd
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