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arcctg2x
Plotting a function graph/ arcctg2x

Graphing y = arcctg2x

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = acot(2*x)
f(x)=acot⁡(2x)f{\left(x \right)} = \operatorname{acot}{\left(2 x \right)} 
f = acot(2*x)
The graph of the function
02468-8-6-4-2-10105-5
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:
acot⁡(2x)=0\operatorname{acot}{\left(2 x \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 acot(2*x).
acot⁡(2⋅0)\operatorname{acot}{\left(2 \cdot 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
−24x2+1=0- \frac{2}{4 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
16x(4x2+1)2=0\frac{16 x}{\left(4 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→−∞acot⁡(2x)=0\lim_{x \to -\infty} \operatorname{acot}{\left(2 x \right)} = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞acot⁡(2x)=0\lim_{x \to \infty} \operatorname{acot}{\left(2 x \right)} = 0
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=0y = 0
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of acot(2*x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(acot⁡(2x)x)=0\lim_{x \to -\infty}\left(\frac{\operatorname{acot}{\left(2 x \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(acot⁡(2x)x)=0\lim_{x \to \infty}\left(\frac{\operatorname{acot}{\left(2 x \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:
acot⁡(2x)=−acot⁡(2x)\operatorname{acot}{\left(2 x \right)} = - \operatorname{acot}{\left(2 x \right)}
- No
acot⁡(2x)=acot⁡(2x)\operatorname{acot}{\left(2 x \right)} = \operatorname{acot}{\left(2 x \right)}
- Yes
so, the function
is
odd
The graph
Graphing y = arcctg2x
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