Mister Exam

Other calculators

Plotting a function graph/ f(x)=|x+2|/arctg(x+2)

Graphing y = f(x)=|x+2|/arctg(x+2)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
         |x + 2|  
f(x) = -----------
       atan(x + 2)
f(x)=∣x+2∣atan⁡(x+2)f{\left(x \right)} = \frac{\left|{x + 2}\right|}{\operatorname{atan}{\left(x + 2 \right)}} 
f = |x + 2|/atan(x + 2)
The graph of the function
02468-8-6-4-2-1010-1010
The domain of the function
The points at which the function is not precisely defined:
x1=−2x_{1} = -2
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:
∣x+2∣atan⁡(x+2)=0\frac{\left|{x + 2}\right|}{\operatorname{atan}{\left(x + 2 \right)}} = 0
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to |x + 2|/atan(x + 2).
∣2∣atan⁡(2)\frac{\left|{2}\right|}{\operatorname{atan}{\left(2 \right)}}
The result:
f(0)=2atan⁡(2)f{\left(0 \right)} = \frac{2}{\operatorname{atan}{\left(2 \right)}}
The point:
(0, 2/atan(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
sign⁡(x+2)atan⁡(x+2)−∣x+2∣((x+2)2+1)atan⁡2(x+2)=0\frac{\operatorname{sign}{\left(x + 2 \right)}}{\operatorname{atan}{\left(x + 2 \right)}} - \frac{\left|{x + 2}\right|}{\left(\left(x + 2\right)^{2} + 1\right) \operatorname{atan}^{2}{\left(x + 2 \right)}} = 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(δ(x+2)−sign⁡(x+2)((x+2)2+1)atan⁡(x+2)+(x+2+1atan⁡(x+2))∣x+2∣((x+2)2+1)2atan⁡(x+2))atan⁡(x+2)=0\frac{2 \left(\delta\left(x + 2\right) - \frac{\operatorname{sign}{\left(x + 2 \right)}}{\left(\left(x + 2\right)^{2} + 1\right) \operatorname{atan}{\left(x + 2 \right)}} + \frac{\left(x + 2 + \frac{1}{\operatorname{atan}{\left(x + 2 \right)}}\right) \left|{x + 2}\right|}{\left(\left(x + 2\right)^{2} + 1\right)^{2} \operatorname{atan}{\left(x + 2 \right)}}\right)}{\operatorname{atan}{\left(x + 2 \right)}} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
x1=−2x_{1} = -2
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(∣x+2∣atan⁡(x+2))=−∞\lim_{x \to -\infty}\left(\frac{\left|{x + 2}\right|}{\operatorname{atan}{\left(x + 2 \right)}}\right) = -\infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞(∣x+2∣atan⁡(x+2))=∞\lim_{x \to \infty}\left(\frac{\left|{x + 2}\right|}{\operatorname{atan}{\left(x + 2 \right)}}\right) = \infty
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of |x + 2|/atan(x + 2), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(∣x+2∣xatan⁡(x+2))=2π\lim_{x \to -\infty}\left(\frac{\left|{x + 2}\right|}{x \operatorname{atan}{\left(x + 2 \right)}}\right) = \frac{2}{\pi}
Let's take the limit
so,
inclined asymptote equation on the left:
y=2xπy = \frac{2 x}{\pi}
lim⁡x→∞(∣x+2∣xatan⁡(x+2))=2π\lim_{x \to \infty}\left(\frac{\left|{x + 2}\right|}{x \operatorname{atan}{\left(x + 2 \right)}}\right) = \frac{2}{\pi}
Let's take the limit
so,
inclined asymptote equation on the right:
y=2xπy = \frac{2 x}{\pi}
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:
∣x+2∣atan⁡(x+2)=−∣x−2∣atan⁡(x−2)\frac{\left|{x + 2}\right|}{\operatorname{atan}{\left(x + 2 \right)}} = - \frac{\left|{x - 2}\right|}{\operatorname{atan}{\left(x - 2 \right)}}
- No
∣x+2∣atan⁡(x+2)=∣x−2∣atan⁡(x−2)\frac{\left|{x + 2}\right|}{\operatorname{atan}{\left(x + 2 \right)}} = \frac{\left|{x - 2}\right|}{\operatorname{atan}{\left(x - 2 \right)}}
- No
so, the function
not is
neither even, nor odd
    © Mister Exam – Calculator