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Plotting a function graph/ -arctg(x+1)

Graphing y = -arctg(x+1)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
f(x) = -atan(x + 1)
f(x)=−atan⁡(x+1)f{\left(x \right)} = - \operatorname{atan}{\left(x + 1 \right)} 
f = -atan(x + 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:
−atan⁡(x+1)=0- \operatorname{atan}{\left(x + 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 -atan(x + 1).
−atan⁡(1)- \operatorname{atan}{\left(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
−1(x+1)2+1=0- \frac{1}{\left(x + 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
2(x+1)((x+1)2+1)2=0\frac{2 \left(x + 1\right)}{\left(\left(x + 1\right)^{2} + 1\right)^{2}} = 0
Solve this equation
The roots of this equation
x1=−1x_{1} = -1

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
[−1,∞)\left[-1, \infty\right)
Convex at the intervals
(−∞,−1]\left(-\infty, -1\right]
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(−atan⁡(x+1))=π2\lim_{x \to -\infty}\left(- \operatorname{atan}{\left(x + 1 \right)}\right) = \frac{\pi}{2}
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=π2y = \frac{\pi}{2}
lim⁡x→∞(−atan⁡(x+1))=−π2\lim_{x \to \infty}\left(- \operatorname{atan}{\left(x + 1 \right)}\right) = - \frac{\pi}{2}
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=−π2y = - \frac{\pi}{2}
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of -atan(x + 1), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(−atan⁡(x+1)x)=0\lim_{x \to -\infty}\left(- \frac{\operatorname{atan}{\left(x + 1 \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(−atan⁡(x+1)x)=0\lim_{x \to \infty}\left(- \frac{\operatorname{atan}{\left(x + 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:
−atan⁡(x+1)=atan⁡(x−1)- \operatorname{atan}{\left(x + 1 \right)} = \operatorname{atan}{\left(x - 1 \right)}
- No
−atan⁡(x+1)=−atan⁡(x−1)- \operatorname{atan}{\left(x + 1 \right)} = - \operatorname{atan}{\left(x - 1 \right)}
- No
so, the function
not is
neither even, nor odd
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