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

Graphing y = arctg((x-1)/x)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
           /x - 1\
f(x) = atan|-----|
           \  x  /
f(x)=atan⁡(x−1x)f{\left(x \right)} = \operatorname{atan}{\left(\frac{x - 1}{x} \right)} 
f = atan((x - 1*1)/x)
The graph of the function
02468-8-6-4-2-10105-5
The domain of the function
The points at which the function is not precisely defined:
x1=0x_{1} = 0
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−1x)=0\operatorname{atan}{\left(\frac{x - 1}{x} \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*1)/x).
atan⁡((−1)1+00)\operatorname{atan}{\left(\frac{\left(-1\right) 1 + 0}{0} \right)}
The result:
f(0)=⟨−π2,π2⟩f{\left(0 \right)} = \left\langle - \frac{\pi}{2}, \frac{\pi}{2}\right\rangle
The point:
(0, AccumBounds(-pi/2, 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
1x−x−1x21+(x−1)2x2=0\frac{\frac{1}{x} - \frac{x - 1}{x^{2}}}{1 + \frac{\left(x - 1\right)^{2}}{x^{2}}} = 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⋅(1−x−1x)(1+(1−x−1x)(x−1)x(1+(x−1)2x2))x2⋅(1+(x−1)2x2)=0- \frac{2 \cdot \left(1 - \frac{x - 1}{x}\right) \left(1 + \frac{\left(1 - \frac{x - 1}{x}\right) \left(x - 1\right)}{x \left(1 + \frac{\left(x - 1\right)^{2}}{x^{2}}\right)}\right)}{x^{2} \cdot \left(1 + \frac{\left(x - 1\right)^{2}}{x^{2}}\right)} = 0
Solve this equation
The roots of this equation
x1=12x_{1} = \frac{1}{2}
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=0x_{1} = 0

lim⁡x→0−(−2⋅(1−x−1x)(1+(1−x−1x)(x−1)x(1+(x−1)2x2))x2⋅(1+(x−1)2x2))=2\lim_{x \to 0^-}\left(- \frac{2 \cdot \left(1 - \frac{x - 1}{x}\right) \left(1 + \frac{\left(1 - \frac{x - 1}{x}\right) \left(x - 1\right)}{x \left(1 + \frac{\left(x - 1\right)^{2}}{x^{2}}\right)}\right)}{x^{2} \cdot \left(1 + \frac{\left(x - 1\right)^{2}}{x^{2}}\right)}\right) = 2
Let's take the limit
lim⁡x→0+(−2⋅(1−x−1x)(1+(1−x−1x)(x−1)x(1+(x−1)2x2))x2⋅(1+(x−1)2x2))=2\lim_{x \to 0^+}\left(- \frac{2 \cdot \left(1 - \frac{x - 1}{x}\right) \left(1 + \frac{\left(1 - \frac{x - 1}{x}\right) \left(x - 1\right)}{x \left(1 + \frac{\left(x - 1\right)^{2}}{x^{2}}\right)}\right)}{x^{2} \cdot \left(1 + \frac{\left(x - 1\right)^{2}}{x^{2}}\right)}\right) = 2
Let's take the limit
- 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
(−∞,12]\left(-\infty, \frac{1}{2}\right]
Convex at the intervals
[12,∞)\left[\frac{1}{2}, \infty\right)
Vertical asymptotes
Have:
x1=0x_{1} = 0
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−1x)=lim⁡x→−∞atan⁡(x−1x)\lim_{x \to -\infty} \operatorname{atan}{\left(\frac{x - 1}{x} \right)} = \lim_{x \to -\infty} \operatorname{atan}{\left(\frac{x - 1}{x} \right)}
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=lim⁡x→−∞atan⁡(x−1x)y = \lim_{x \to -\infty} \operatorname{atan}{\left(\frac{x - 1}{x} \right)}
lim⁡x→∞atan⁡(x−1x)=lim⁡x→∞atan⁡(x−1x)\lim_{x \to \infty} \operatorname{atan}{\left(\frac{x - 1}{x} \right)} = \lim_{x \to \infty} \operatorname{atan}{\left(\frac{x - 1}{x} \right)}
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=lim⁡x→∞atan⁡(x−1x)y = \lim_{x \to \infty} \operatorname{atan}{\left(\frac{x - 1}{x} \right)}
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of atan((x - 1*1)/x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(atan⁡(x−1x)x)=lim⁡x→−∞(atan⁡(x−1x)x)\lim_{x \to -\infty}\left(\frac{\operatorname{atan}{\left(\frac{x - 1}{x} \right)}}{x}\right) = \lim_{x \to -\infty}\left(\frac{\operatorname{atan}{\left(\frac{x - 1}{x} \right)}}{x}\right)
Let's take the limit
so,
inclined asymptote equation on the left:
y=xlim⁡x→−∞(atan⁡(x−1x)x)y = x \lim_{x \to -\infty}\left(\frac{\operatorname{atan}{\left(\frac{x - 1}{x} \right)}}{x}\right)
lim⁡x→∞(atan⁡(x−1x)x)=lim⁡x→∞(atan⁡(x−1x)x)\lim_{x \to \infty}\left(\frac{\operatorname{atan}{\left(\frac{x - 1}{x} \right)}}{x}\right) = \lim_{x \to \infty}\left(\frac{\operatorname{atan}{\left(\frac{x - 1}{x} \right)}}{x}\right)
Let's take the limit
so,
inclined asymptote equation on the right:
y=xlim⁡x→∞(atan⁡(x−1x)x)y = x \lim_{x \to \infty}\left(\frac{\operatorname{atan}{\left(\frac{x - 1}{x} \right)}}{x}\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:
atan⁡(x−1x)=−atan⁡(−x−1x)\operatorname{atan}{\left(\frac{x - 1}{x} \right)} = - \operatorname{atan}{\left(\frac{- x - 1}{x} \right)}
- No
atan⁡(x−1x)=atan⁡(−x−1x)\operatorname{atan}{\left(\frac{x - 1}{x} \right)} = \operatorname{atan}{\left(\frac{- x - 1}{x} \right)}
- No
so, the function
not is
neither even, nor odd
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