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Plotting a function graph/ arcsin(sqrt(2x-1))

Graphing y = arcsin(sqrt(2x-1))

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
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f(x) = asin\\/ 2*x - 1 /
f(x)=asin⁡(2x−1)f{\left(x \right)} = \operatorname{asin}{\left(\sqrt{2 x - 1} \right)} 
f = asin(sqrt(2*x - 1))
The graph of the function
02468-8-6-4-2-101002
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⁡(2x−1)=0\operatorname{asin}{\left(\sqrt{2 x - 1} \right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=12x_{1} = \frac{1}{2}
Numerical solution
x1=0.5x_{1} = 0.5
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to asin(sqrt(2*x - 1)).
asin⁡(−1+0⋅2)\operatorname{asin}{\left(\sqrt{-1 + 0 \cdot 2} \right)}
The result:
f(0)=ilog⁡(1+2)f{\left(0 \right)} = i \log{\left(1 + \sqrt{2} \right)}
The point:
(0, i*log(1 + sqrt(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
12−2x2x−1=0\frac{1}{\sqrt{2 - 2 x} \sqrt{2 x - 1}} = 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(−22x−1+11−x)41−x2x−1=0\frac{\sqrt{2} \left(- \frac{2}{2 x - 1} + \frac{1}{1 - x}\right)}{4 \sqrt{1 - x} \sqrt{2 x - 1}} = 0
Solve this equation
The roots of this equation
x1=34x_{1} = \frac{3}{4}

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
[34,∞)\left[\frac{3}{4}, \infty\right)
Convex at the intervals
(−∞,34]\left(-\infty, \frac{3}{4}\right]
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞asin⁡(2x−1)=∞i\lim_{x \to -\infty} \operatorname{asin}{\left(\sqrt{2 x - 1} \right)} = \infty i
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞asin⁡(2x−1)=−∞i\lim_{x \to \infty} \operatorname{asin}{\left(\sqrt{2 x - 1} \right)} = - \infty i
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 asin(sqrt(2*x - 1)), 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→−∞(asin⁡(2x−1)x)y = x \lim_{x \to -\infty}\left(\frac{\operatorname{asin}{\left(\sqrt{2 x - 1} \right)}}{x}\right)
True

Let's take the limit
so,
inclined asymptote equation on the right:
y=xlim⁡x→∞(asin⁡(2x−1)x)y = x \lim_{x \to \infty}\left(\frac{\operatorname{asin}{\left(\sqrt{2 x - 1} \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:
asin⁡(2x−1)=asin⁡(−2x−1)\operatorname{asin}{\left(\sqrt{2 x - 1} \right)} = \operatorname{asin}{\left(\sqrt{- 2 x - 1} \right)}
- No
asin⁡(2x−1)=−asin⁡(−2x−1)\operatorname{asin}{\left(\sqrt{2 x - 1} \right)} = - \operatorname{asin}{\left(\sqrt{- 2 x - 1} \right)}
- No
so, the function
not is
neither even, nor odd
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