Mister Exam

Other calculators

Plotting a function graph/ arctg((x+2)/(x-3))

Graphing y = arctg((x+2)/(x-3))

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
           /x + 2\
f(x) = atan|-----|
           \x - 3/
f(x)=atan⁡(x+2x−3)f{\left(x \right)} = \operatorname{atan}{\left(\frac{x + 2}{x - 3} \right)} 
f = atan((x + 2)/(x - 3))
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=3x_{1} = 3
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+2x−3)=0\operatorname{atan}{\left(\frac{x + 2}{x - 3} \right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−2x_{1} = -2
Numerical solution
x1=−2x_{1} = -2
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to atan((x + 2)/(x - 3)).
atan⁡(2−3)\operatorname{atan}{\left(\frac{2}{-3} \right)}
The result:
f(0)=−atan⁡(23)f{\left(0 \right)} = - \operatorname{atan}{\left(\frac{2}{3} \right)}
The point:
(0, -atan(2/3))
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−3−x+2(x−3)21+(x+2)2(x−3)2=0\frac{\frac{1}{x - 3} - \frac{x + 2}{\left(x - 3\right)^{2}}}{1 + \frac{\left(x + 2\right)^{2}}{\left(x - 3\right)^{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+2x−3)(1+(1−x+2x−3)(x+2)(1+(x+2)2(x−3)2)(x−3))(1+(x+2)2(x−3)2)(x−3)2=0- \frac{2 \left(1 - \frac{x + 2}{x - 3}\right) \left(1 + \frac{\left(1 - \frac{x + 2}{x - 3}\right) \left(x + 2\right)}{\left(1 + \frac{\left(x + 2\right)^{2}}{\left(x - 3\right)^{2}}\right) \left(x - 3\right)}\right)}{\left(1 + \frac{\left(x + 2\right)^{2}}{\left(x - 3\right)^{2}}\right) \left(x - 3\right)^{2}} = 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=3x_{1} = 3

lim⁡x→3−(−2(1−x+2x−3)(1+(1−x+2x−3)(x+2)(1+(x+2)2(x−3)2)(x−3))(1+(x+2)2(x−3)2)(x−3)2)=0.08\lim_{x \to 3^-}\left(- \frac{2 \left(1 - \frac{x + 2}{x - 3}\right) \left(1 + \frac{\left(1 - \frac{x + 2}{x - 3}\right) \left(x + 2\right)}{\left(1 + \frac{\left(x + 2\right)^{2}}{\left(x - 3\right)^{2}}\right) \left(x - 3\right)}\right)}{\left(1 + \frac{\left(x + 2\right)^{2}}{\left(x - 3\right)^{2}}\right) \left(x - 3\right)^{2}}\right) = 0.08
lim⁡x→3+(−2(1−x+2x−3)(1+(1−x+2x−3)(x+2)(1+(x+2)2(x−3)2)(x−3))(1+(x+2)2(x−3)2)(x−3)2)=0.08\lim_{x \to 3^+}\left(- \frac{2 \left(1 - \frac{x + 2}{x - 3}\right) \left(1 + \frac{\left(1 - \frac{x + 2}{x - 3}\right) \left(x + 2\right)}{\left(1 + \frac{\left(x + 2\right)^{2}}{\left(x - 3\right)^{2}}\right) \left(x - 3\right)}\right)}{\left(1 + \frac{\left(x + 2\right)^{2}}{\left(x - 3\right)^{2}}\right) \left(x - 3\right)^{2}}\right) = 0.08
- 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[\frac{1}{2}, \infty\right)
Convex at the intervals
(−∞,12]\left(-\infty, \frac{1}{2}\right]
Vertical asymptotes
Have:
x1=3x_{1} = 3
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+2x−3)=π4\lim_{x \to -\infty} \operatorname{atan}{\left(\frac{x + 2}{x - 3} \right)} = \frac{\pi}{4}
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=π4y = \frac{\pi}{4}
lim⁡x→∞atan⁡(x+2x−3)=π4\lim_{x \to \infty} \operatorname{atan}{\left(\frac{x + 2}{x - 3} \right)} = \frac{\pi}{4}
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=π4y = \frac{\pi}{4}
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of atan((x + 2)/(x - 3)), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(atan⁡(x+2x−3)x)=0\lim_{x \to -\infty}\left(\frac{\operatorname{atan}{\left(\frac{x + 2}{x - 3} \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(atan⁡(x+2x−3)x)=0\lim_{x \to \infty}\left(\frac{\operatorname{atan}{\left(\frac{x + 2}{x - 3} \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+2x−3)=atan⁡(2−x−x−3)\operatorname{atan}{\left(\frac{x + 2}{x - 3} \right)} = \operatorname{atan}{\left(\frac{2 - x}{- x - 3} \right)}
- No
atan⁡(x+2x−3)=−atan⁡(2−x−x−3)\operatorname{atan}{\left(\frac{x + 2}{x - 3} \right)} = - \operatorname{atan}{\left(\frac{2 - x}{- x - 3} \right)}
- No
so, the function
not is
neither even, nor odd
    © Mister Exam – Calculator