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Plotting a function graph/ arcctgx-3x

Graphing y = arcctgx-3x

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = acot(x) - 3*x
f(x)=−3x+acot⁡(x)f{\left(x \right)} = - 3 x + \operatorname{acot}{\left(x \right)} 
f = -3*x + acot(x)
The graph of the function
02468-8-6-4-2-1010-5050
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:
−3x+acot⁡(x)=0- 3 x + \operatorname{acot}{\left(x \right)} = 0
Solve this equation
The points of intersection with the axis X:

Numerical solution
x1=0.397486276445476x_{1} = 0.397486276445476
x2=−0.397486276445476x_{2} = -0.397486276445476
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to acot(x) - 3*x.
−0+acot⁡(0)- 0 + \operatorname{acot}{\left(0 \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
−3−1x2+1=0-3 - \frac{1}{x^{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
2x(x2+1)2=0\frac{2 x}{\left(x^{2} + 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→−∞(−3x+acot⁡(x))=∞\lim_{x \to -\infty}\left(- 3 x + \operatorname{acot}{\left(x \right)}\right) = \infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞(−3x+acot⁡(x))=−∞\lim_{x \to \infty}\left(- 3 x + \operatorname{acot}{\left(x \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 acot(x) - 3*x, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(−3x+acot⁡(x)x)=−3\lim_{x \to -\infty}\left(\frac{- 3 x + \operatorname{acot}{\left(x \right)}}{x}\right) = -3
Let's take the limit
so,
inclined asymptote equation on the left:
y=−3xy = - 3 x
lim⁡x→∞(−3x+acot⁡(x)x)=−3\lim_{x \to \infty}\left(\frac{- 3 x + \operatorname{acot}{\left(x \right)}}{x}\right) = -3
Let's take the limit
so,
inclined asymptote equation on the right:
y=−3xy = - 3 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:
−3x+acot⁡(x)=3x−acot⁡(x)- 3 x + \operatorname{acot}{\left(x \right)} = 3 x - \operatorname{acot}{\left(x \right)}
- No
−3x+acot⁡(x)=−3x+acot⁡(x)- 3 x + \operatorname{acot}{\left(x \right)} = - 3 x + \operatorname{acot}{\left(x \right)}
- No
so, the function
not is
neither even, nor odd
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