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  • Graphing y =:
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  • x^2-|x-x^2|
  • 3x^2-4x
  • 3x^2-2x
  • 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))

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 /
$$f{\left(x \right)} = \operatorname{acos}{\left(\frac{2 x}{x^{2} + 1} \right)}$$
f = acos((2*x)/(x^2 + 1))
The graph of the function
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:
$$\operatorname{acos}{\left(\frac{2 x}{x^{2} + 1} \right)} = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = 1$$
Numerical solution
$$x_{1} = 1$$
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)).
$$\operatorname{acos}{\left(\frac{0 \cdot 2}{0^{2} + 1} \right)}$$
The result:
$$f{\left(0 \right)} = \frac{\pi}{2}$$
The point:
(0, pi/2)
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$- \frac{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + \frac{2}{x^{2} + 1}}{\sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$- \frac{4 x \left(\frac{4 x^{2}}{x^{2} + 1} - 3 + \frac{2 \left(\frac{2 x^{2}}{x^{2} + 1} - 1\right)^{2}}{\left(x^{2} + 1\right) \left(- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1\right)}\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0$$

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
$$\left[0, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 0\right]$$
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:
$$\operatorname{acos}{\left(\frac{2 x}{x^{2} + 1} \right)} = \operatorname{acos}{\left(- \frac{2 x}{x^{2} + 1} \right)}$$
- No
$$\operatorname{acos}{\left(\frac{2 x}{x^{2} + 1} \right)} = - \operatorname{acos}{\left(- \frac{2 x}{x^{2} + 1} \right)}$$
- No
so, the function
not is
neither even, nor odd