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  • Graphing y =:
  • -x^3+3x^2-2
  • x(2-x)^2
  • -x^2+5x+4
  • 3x^2-4x

  • Identical expressions

  • arccos(two x/(one +x^2))
  • arc co sinus of e of (2x divide by (1 plus x squared ))
  • arc co sinus of e of (two x divide by (one plus x squared ))
  • arccos(2x/(1+x2))
  • arccos2x/1+x2
  • arccos(2x/(1+x²))
  • arccos(2x/(1+x to the power of 2))
  • arccos2x/1+x^2
  • arccos(2x divide by (1+x^2))

  • Similar expressions

  • arccos(2x/(1-x^2))
Plotting a function graph/ arccos(2x/(1+x^2))

Graphing y = arccos(2x/(1+x^2))

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
           / 2*x  \
f(x) = acos|------|
           |     2|
           \1 + x /
f(x)=acos⁡(2xx2+1)f{\left(x \right)} = \operatorname{acos}{\left(\frac{2 x}{x^{2} + 1} \right)} 
f = acos((2*x)/(x^2 + 1))
The graph of the function
-1.0-0.8-0.6-0.4-0.21.00.00.20.40.60.805
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:
acos⁡(2xx2+1)=0\operatorname{acos}{\left(\frac{2 x}{x^{2} + 1} \right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=1x_{1} = 1
Numerical solution
x1=1x_{1} = 1
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to acos((2*x)/(1 + x^2)).
acos⁡(0⋅202+1)\operatorname{acos}{\left(\frac{0 \cdot 2}{0^{2} + 1} \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
−−4x2(x2+1)2+2x2+1−4x2(x2+1)2+1=0- \frac{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + \frac{2}{x^{2} + 1}}{\sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\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
−4x(4x2x2+1−3+2(2x2x2+1−1)2(x2+1)(−4x2(x2+1)2+1))(x2+1)2−4x2(x2+1)2+1=0- \frac{4 x \left(\frac{4 x^{2}}{x^{2} + 1} - 3 + \frac{2 \left(\frac{2 x^{2}}{x^{2} + 1} - 1\right)^{2}}{\left(x^{2} + 1\right) \left(- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1\right)}\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = 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]
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:
acos⁡(2xx2+1)=acos⁡(−2xx2+1)\operatorname{acos}{\left(\frac{2 x}{x^{2} + 1} \right)} = \operatorname{acos}{\left(- \frac{2 x}{x^{2} + 1} \right)}
- No
acos⁡(2xx2+1)=−acos⁡(−2xx2+1)\operatorname{acos}{\left(\frac{2 x}{x^{2} + 1} \right)} = - \operatorname{acos}{\left(- \frac{2 x}{x^{2} + 1} \right)}
- No
so, the function
not is
neither even, nor odd
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