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Plotting a function graph/ arcsin(x-3)/2-(ln(4-x))

Graphing y = arcsin(x-3)/2-(ln(4-x))

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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       asin(x - 3)             
f(x) = ----------- - log(4 - x)
            2
f(x)=−log⁡(4−x)+asin⁡(x−3)2f{\left(x \right)} = - \log{\left(4 - x \right)} + \frac{\operatorname{asin}{\left(x - 3 \right)}}{2} 
f = -log(4 - x) + asin(x - 3)/2
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:
−log⁡(4−x)+asin⁡(x−3)2=0- \log{\left(4 - x \right)} + \frac{\operatorname{asin}{\left(x - 3 \right)}}{2} = 0
Solve this equation
The points of intersection with the axis X:

Numerical solution
x1=3x_{1} = 3
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to asin(x - 3)/2 - log(4 - x).
−log⁡(4−0)+asin⁡(−3)2- \log{\left(4 - 0 \right)} + \frac{\operatorname{asin}{\left(-3 \right)}}{2}
The result:
f(0)=−log⁡(4)−asin⁡(3)2f{\left(0 \right)} = - \log{\left(4 \right)} - \frac{\operatorname{asin}{\left(3 \right)}}{2}
The point:
(0, -log(4) - asin(3)/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
14−x+121−(x−3)2=0\frac{1}{4 - x} + \frac{1}{2 \sqrt{1 - \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
1(x−4)2+x−32(1−(x−3)2)32=0\frac{1}{\left(x - 4\right)^{2}} + \frac{x - 3}{2 \left(1 - \left(x - 3\right)^{2}\right)^{\frac{3}{2}}} = 0
Solve this equation
The roots of this equation
x1=−592252893375+23212253+2893375+23212253+3415x_{1} = - \frac{59}{225 \sqrt[3]{\frac{289}{3375} + \frac{2 \sqrt{321}}{225}}} + \sqrt[3]{\frac{289}{3375} + \frac{2 \sqrt{321}}{225}} + \frac{34}{15}

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
[−592252893375+23212253+2893375+23212253+3415,∞)\left[- \frac{59}{225 \sqrt[3]{\frac{289}{3375} + \frac{2 \sqrt{321}}{225}}} + \sqrt[3]{\frac{289}{3375} + \frac{2 \sqrt{321}}{225}} + \frac{34}{15}, \infty\right)
Convex at the intervals
(−∞,−592252893375+23212253+2893375+23212253+3415]\left(-\infty, - \frac{59}{225 \sqrt[3]{\frac{289}{3375} + \frac{2 \sqrt{321}}{225}}} + \sqrt[3]{\frac{289}{3375} + \frac{2 \sqrt{321}}{225}} + \frac{34}{15}\right]
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
False

Let's take the limit
so,
equation of the horizontal asymptote on the left:
False

False

Let's take the limit
so,
equation of the horizontal asymptote on the right:
False
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of asin(x - 3)/2 - log(4 - x), divided by x at x->+oo and x ->-oo
True

Let's take the limit
so,
inclined asymptote equation on the left:
y=xlim⁡x→−∞(−log⁡(4−x)+asin⁡(x−3)2x)y = x \lim_{x \to -\infty}\left(\frac{- \log{\left(4 - x \right)} + \frac{\operatorname{asin}{\left(x - 3 \right)}}{2}}{x}\right)
lim⁡x→∞(−log⁡(4−x)+asin⁡(x−3)2x)=0\lim_{x \to \infty}\left(\frac{- \log{\left(4 - x \right)} + \frac{\operatorname{asin}{\left(x - 3 \right)}}{2}}{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:
−log⁡(4−x)+asin⁡(x−3)2=−log⁡(x+4)−asin⁡(x+3)2- \log{\left(4 - x \right)} + \frac{\operatorname{asin}{\left(x - 3 \right)}}{2} = - \log{\left(x + 4 \right)} - \frac{\operatorname{asin}{\left(x + 3 \right)}}{2}
- No
−log⁡(4−x)+asin⁡(x−3)2=log⁡(x+4)+asin⁡(x+3)2- \log{\left(4 - x \right)} + \frac{\operatorname{asin}{\left(x - 3 \right)}}{2} = \log{\left(x + 4 \right)} + \frac{\operatorname{asin}{\left(x + 3 \right)}}{2}
- No
so, the function
not is
neither even, nor odd
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