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Plotting a function graph/ arcsin((x+5)/(2-x))

Graphing y = arcsin((x+5)/(2-x))

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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           /x + 5\
f(x) = asin|-----|
           \2 - x/
f(x)=asin⁡(x+52−x)f{\left(x \right)} = \operatorname{asin}{\left(\frac{x + 5}{2 - x} \right)} 
f = asin((x + 5)/(2 - x))
The graph of the function
02468-8-6-4-2-10102-2
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:
asin⁡(x+52−x)=0\operatorname{asin}{\left(\frac{x + 5}{2 - x} \right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−5x_{1} = -5
Numerical solution
x1=−5x_{1} = -5
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to asin((x + 5)/(2 - x)).
asin⁡(52−0)\operatorname{asin}{\left(\frac{5}{2 - 0} \right)}
The result:
f(0)=asin⁡(52)f{\left(0 \right)} = \operatorname{asin}{\left(\frac{5}{2} \right)}
The point:
(0, asin(5/2))
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
12−x+x+5(2−x)21−(x+5)2(2−x)2=0\frac{\frac{1}{2 - x} + \frac{x + 5}{\left(2 - x\right)^{2}}}{\sqrt{1 - \frac{\left(x + 5\right)^{2}}{\left(2 - x\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
(1−x+5x−2)(2−(1−x+5x−2)(x+5)(1−(x+5)2(x−2)2)(x−2))1−(x+5)2(x−2)2(x−2)2=0\frac{\left(1 - \frac{x + 5}{x - 2}\right) \left(2 - \frac{\left(1 - \frac{x + 5}{x - 2}\right) \left(x + 5\right)}{\left(1 - \frac{\left(x + 5\right)^{2}}{\left(x - 2\right)^{2}}\right) \left(x - 2\right)}\right)}{\sqrt{1 - \frac{\left(x + 5\right)^{2}}{\left(x - 2\right)^{2}}} \left(x - 2\right)^{2}} = 0
Solve this equation
The roots of this equation
x1=−13x_{1} = - \frac{1}{3}
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=2x_{1} = 2

lim⁡x→2−((1−x+5x−2)(2−(1−x+5x−2)(x+5)(1−(x+5)2(x−2)2)(x−2))1−(x+5)2(x−2)2(x−2)2)=−∞i\lim_{x \to 2^-}\left(\frac{\left(1 - \frac{x + 5}{x - 2}\right) \left(2 - \frac{\left(1 - \frac{x + 5}{x - 2}\right) \left(x + 5\right)}{\left(1 - \frac{\left(x + 5\right)^{2}}{\left(x - 2\right)^{2}}\right) \left(x - 2\right)}\right)}{\sqrt{1 - \frac{\left(x + 5\right)^{2}}{\left(x - 2\right)^{2}}} \left(x - 2\right)^{2}}\right) = - \infty i
lim⁡x→2+((1−x+5x−2)(2−(1−x+5x−2)(x+5)(1−(x+5)2(x−2)2)(x−2))1−(x+5)2(x−2)2(x−2)2)=∞i\lim_{x \to 2^+}\left(\frac{\left(1 - \frac{x + 5}{x - 2}\right) \left(2 - \frac{\left(1 - \frac{x + 5}{x - 2}\right) \left(x + 5\right)}{\left(1 - \frac{\left(x + 5\right)^{2}}{\left(x - 2\right)^{2}}\right) \left(x - 2\right)}\right)}{\sqrt{1 - \frac{\left(x + 5\right)^{2}}{\left(x - 2\right)^{2}}} \left(x - 2\right)^{2}}\right) = \infty i
- the limits are not equal, so
x1=2x_{1} = 2
- is an inflection 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:
Have no bends at the whole real axis
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→−∞asin⁡(x+52−x)=−π2\lim_{x \to -\infty} \operatorname{asin}{\left(\frac{x + 5}{2 - x} \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→∞asin⁡(x+52−x)=−π2\lim_{x \to \infty} \operatorname{asin}{\left(\frac{x + 5}{2 - x} \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 asin((x + 5)/(2 - x)), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(asin⁡(x+52−x)x)=0\lim_{x \to -\infty}\left(\frac{\operatorname{asin}{\left(\frac{x + 5}{2 - x} \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(asin⁡(x+52−x)x)=0\lim_{x \to \infty}\left(\frac{\operatorname{asin}{\left(\frac{x + 5}{2 - x} \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:
asin⁡(x+52−x)=asin⁡(5−xx+2)\operatorname{asin}{\left(\frac{x + 5}{2 - x} \right)} = \operatorname{asin}{\left(\frac{5 - x}{x + 2} \right)}
- No
asin⁡(x+52−x)=−asin⁡(5−xx+2)\operatorname{asin}{\left(\frac{x + 5}{2 - x} \right)} = - \operatorname{asin}{\left(\frac{5 - x}{x + 2} \right)}
- No
so, the function
not is
neither even, nor odd
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