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  • Graphing y =:
  • x^5-x
  • x^3-6x^2-15x-8
  • -x^2+9
  • -x^2-6*x-5

  • Identical expressions

  • -arctg((4x)/(one -x^ two))
  • minus arctg((4x) divide by (1 minus x squared ))
  • minus arctg((4x) divide by (one minus x to the power of two))
  • -arctg((4x)/(1-x2))
  • -arctg4x/1-x2
  • -arctg((4x)/(1-x²))
  • -arctg((4x)/(1-x to the power of 2))
  • -arctg4x/1-x^2
  • -arctg((4x) divide by (1-x^2))

  • Similar expressions

  • arctg((4x)/(1-x^2))
  • -arctg((4x)/(1+x^2))
Plotting a function graph/ -arctg((4x)/(1-x^2))

Graphing y = -arctg((4x)/(1-x^2))

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
            / 4*x  \
f(x) = -atan|------|
            |     2|
            \1 - x /
f(x)=−atan⁡(4x1−x2)f{\left(x \right)} = - \operatorname{atan}{\left(\frac{4 x}{1 - x^{2}} \right)} 
f = -atan((4*x)/(1 - x^2))
The graph of the function
0.04.00.51.01.52.02.53.03.55-5
The domain of the function
The points at which the function is not precisely defined:
x1=−1x_{1} = -1
x2=1x_{2} = 1
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:
−atan⁡(4x1−x2)=0- \operatorname{atan}{\left(\frac{4 x}{1 - x^{2}} \right)} = 0
Solve this equation
The points of intersection with the axis X:

Numerical solution
x1=0x_{1} = 0
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to -atan((4*x)/(1 - x^2)).
−atan⁡(0⋅41−02)- \operatorname{atan}{\left(\frac{0 \cdot 4}{1 - 0^{2}} \right)}
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
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
−8x2(1−x2)2+41−x216x2(1−x2)2+1=0- \frac{\frac{8 x^{2}}{\left(1 - x^{2}\right)^{2}} + \frac{4}{1 - x^{2}}}{\frac{16 x^{2}}{\left(1 - x^{2}\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
−8x(−4x2x2−1+3+16(2x2x2−1−1)2(x2−1)(16x2(x2−1)2+1))(x2−1)2(16x2(x2−1)2+1)=0- \frac{8 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3 + \frac{16 \left(\frac{2 x^{2}}{x^{2} - 1} - 1\right)^{2}}{\left(x^{2} - 1\right) \left(\frac{16 x^{2}}{\left(x^{2} - 1\right)^{2}} + 1\right)}\right)}{\left(x^{2} - 1\right)^{2} \left(\frac{16 x^{2}}{\left(x^{2} - 1\right)^{2}} + 1\right)} = 0
Solve this equation
The roots of this equation
x1=0x_{1} = 0
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
x1=−1x_{1} = -1
x2=1x_{2} = 1

lim⁡x→−1−(−8x(−4x2x2−1+3+16(2x2x2−1−1)2(x2−1)(16x2(x2−1)2+1))(x2−1)2(16x2(x2−1)2+1))=−0.5\lim_{x \to -1^-}\left(- \frac{8 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3 + \frac{16 \left(\frac{2 x^{2}}{x^{2} - 1} - 1\right)^{2}}{\left(x^{2} - 1\right) \left(\frac{16 x^{2}}{\left(x^{2} - 1\right)^{2}} + 1\right)}\right)}{\left(x^{2} - 1\right)^{2} \left(\frac{16 x^{2}}{\left(x^{2} - 1\right)^{2}} + 1\right)}\right) = -0.5
lim⁡x→−1+(−8x(−4x2x2−1+3+16(2x2x2−1−1)2(x2−1)(16x2(x2−1)2+1))(x2−1)2(16x2(x2−1)2+1))=−0.5\lim_{x \to -1^+}\left(- \frac{8 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3 + \frac{16 \left(\frac{2 x^{2}}{x^{2} - 1} - 1\right)^{2}}{\left(x^{2} - 1\right) \left(\frac{16 x^{2}}{\left(x^{2} - 1\right)^{2}} + 1\right)}\right)}{\left(x^{2} - 1\right)^{2} \left(\frac{16 x^{2}}{\left(x^{2} - 1\right)^{2}} + 1\right)}\right) = -0.5
- limits are equal, then skip the corresponding point
lim⁡x→1−(−8x(−4x2x2−1+3+16(2x2x2−1−1)2(x2−1)(16x2(x2−1)2+1))(x2−1)2(16x2(x2−1)2+1))=0.5\lim_{x \to 1^-}\left(- \frac{8 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3 + \frac{16 \left(\frac{2 x^{2}}{x^{2} - 1} - 1\right)^{2}}{\left(x^{2} - 1\right) \left(\frac{16 x^{2}}{\left(x^{2} - 1\right)^{2}} + 1\right)}\right)}{\left(x^{2} - 1\right)^{2} \left(\frac{16 x^{2}}{\left(x^{2} - 1\right)^{2}} + 1\right)}\right) = 0.5
lim⁡x→1+(−8x(−4x2x2−1+3+16(2x2x2−1−1)2(x2−1)(16x2(x2−1)2+1))(x2−1)2(16x2(x2−1)2+1))=0.5\lim_{x \to 1^+}\left(- \frac{8 x \left(- \frac{4 x^{2}}{x^{2} - 1} + 3 + \frac{16 \left(\frac{2 x^{2}}{x^{2} - 1} - 1\right)^{2}}{\left(x^{2} - 1\right) \left(\frac{16 x^{2}}{\left(x^{2} - 1\right)^{2}} + 1\right)}\right)}{\left(x^{2} - 1\right)^{2} \left(\frac{16 x^{2}}{\left(x^{2} - 1\right)^{2}} + 1\right)}\right) = 0.5
- limits are equal, then skip the corresponding point

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]
Vertical asymptotes
Have:
x1=−1x_{1} = -1
x2=1x_{2} = 1
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(−atan⁡(4x1−x2))=0\lim_{x \to -\infty}\left(- \operatorname{atan}{\left(\frac{4 x}{1 - x^{2}} \right)}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(−atan⁡(4x1−x2))=0\lim_{x \to \infty}\left(- \operatorname{atan}{\left(\frac{4 x}{1 - x^{2}} \right)}\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 -atan((4*x)/(1 - x^2)), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(−atan⁡(4x1−x2)x)=0\lim_{x \to -\infty}\left(- \frac{\operatorname{atan}{\left(\frac{4 x}{1 - x^{2}} \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(−atan⁡(4x1−x2)x)=0\lim_{x \to \infty}\left(- \frac{\operatorname{atan}{\left(\frac{4 x}{1 - x^{2}} \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:
−atan⁡(4x1−x2)=atan⁡(4x1−x2)- \operatorname{atan}{\left(\frac{4 x}{1 - x^{2}} \right)} = \operatorname{atan}{\left(\frac{4 x}{1 - x^{2}} \right)}
- No
−atan⁡(4x1−x2)=−atan⁡(4x1−x2)- \operatorname{atan}{\left(\frac{4 x}{1 - x^{2}} \right)} = - \operatorname{atan}{\left(\frac{4 x}{1 - x^{2}} \right)}
- No
so, the function
not is
neither even, nor odd
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