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

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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           /  _________\
f(x) = asin\\/ 2*x - 1 /
$$f{\left(x \right)} = \operatorname{asin}{\left(\sqrt{2 x - 1} \right)}$$
f = asin(sqrt(2*x - 1))
The graph of the function
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:
$$\operatorname{asin}{\left(\sqrt{2 x - 1} \right)} = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = \frac{1}{2}$$
Numerical solution
$$x_{1} = 0.5$$
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)).
$$\operatorname{asin}{\left(\sqrt{-1 + 0 \cdot 2} \right)}$$
The result:
$$f{\left(0 \right)} = i \log{\left(1 + \sqrt{2} \right)}$$
The point:
(0, i*log(1 + sqrt(2)))
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$\frac{1}{\sqrt{2 - 2 x} \sqrt{2 x - 1}} = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$\frac{\sqrt{2} \left(- \frac{2}{2 x - 1} + \frac{1}{1 - x}\right)}{4 \sqrt{1 - x} \sqrt{2 x - 1}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = \frac{3}{4}$$

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

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