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  • Graphing y =:
  • x^4-4x^3+4
  • x^4-4x^2+2
  • x^4-2x^3+4
  • x^4-2x^3+3

  • Identical expressions

  • acot(two *x/ five + one)
  • arcco tangent of gent of (2 multiply by x divide by 5 plus 1)
  • arcco tangent of gent of (two multiply by x divide by five plus one)
  • acot(2x/5+1)
  • acot2x/5+1
  • acot(2*x divide by 5+1)

  • Similar expressions

  • acot(2*x/5-1)
  • arccot(2*x/5+1)
Plotting a function graph/ acot(2*x/5+1)

Graphing y = acot(2*x/5+1)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
           /2*x    \
f(x) = acot|--- + 1|
           \ 5     /
f(x)=acot⁡(2x5+1)f{\left(x \right)} = \operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} 
f = acot((2*x)/5 + 1)
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⁡(2x5+1)=0\operatorname{acot}{\left(\frac{2 x}{5} + 1 \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)/5 + 1).
acot⁡(0⋅25+1)\operatorname{acot}{\left(\frac{0 \cdot 2}{5} + 1 \right)}
The result:
f(0)=π4f{\left(0 \right)} = \frac{\pi}{4}
The point:
(0, pi/4)
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
−25((2x5+1)2+1)=0- \frac{2}{5 \left(\left(\frac{2 x}{5} + 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
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
8(2x+5)125((2x+5)225+1)2=0\frac{8 \left(2 x + 5\right)}{125 \left(\frac{\left(2 x + 5\right)^{2}}{25} + 1\right)^{2}} = 0
Solve this equation
The roots of this equation
x1=−52x_{1} = - \frac{5}{2}

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
[−52,∞)\left[- \frac{5}{2}, \infty\right)
Convex at the intervals
(−∞,−52]\left(-\infty, - \frac{5}{2}\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⁡(2x5+1)=π\lim_{x \to -\infty} \operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = \pi
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=πy = \pi
lim⁡x→∞acot⁡(2x5+1)=0\lim_{x \to \infty} \operatorname{acot}{\left(\frac{2 x}{5} + 1 \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)/5 + 1), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(acot⁡(2x5+1)x)=0\lim_{x \to -\infty}\left(\frac{\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(acot⁡(2x5+1)x)=0\lim_{x \to \infty}\left(\frac{\operatorname{acot}{\left(\frac{2 x}{5} + 1 \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⁡(2x5+1)=−acot⁡(2x5−1)\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = - \operatorname{acot}{\left(\frac{2 x}{5} - 1 \right)}
- No
acot⁡(2x5+1)=acot⁡(2x5−1)\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = \operatorname{acot}{\left(\frac{2 x}{5} - 1 \right)}
- No
so, the function
not is
neither even, nor odd
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