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arcsin(1/x^2)

  • How to use it?

  • Graphing y =:
  • x^3+3x^2-9x+15
  • -x^2+6x-8
  • x^2-6x+10
  • (x^2+5)/(x-2)

  • Identical expressions

  • arcsin(one /x^ two)
  • arc sinus of (1 divide by x squared )
  • arc sinus of (one divide by x to the power of two)
  • arcsin(1/x2)
  • arcsin1/x2
  • arcsin(1/x²)
  • arcsin(1/x to the power of 2)
  • arcsin1/x^2
  • arcsin(1 divide by x^2)
Plotting a function graph/ arcsin(1/x^2)

Graphing y = arcsin(1/x^2)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
           /  1 \
f(x) = asin|1*--|
           |   2|
           \  x /
f(x)=asin⁡(1⋅1x2)f{\left(x \right)} = \operatorname{asin}{\left(1 \cdot \frac{1}{x^{2}} \right)} 
f = asin(1/x^2)
The graph of the function
02468-8-6-4-2-101002
The domain of the function
The points at which the function is not precisely defined:
x1=0x_{1} = 0
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⁡(1⋅1x2)=0\operatorname{asin}{\left(1 \cdot \frac{1}{x^{2}} \right)} = 0
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to asin(1/x^2).
asin⁡(1⋅102)\operatorname{asin}{\left(1 \cdot \frac{1}{0^{2}} \right)}
The result:
f(0)=∞~f{\left(0 \right)} = \tilde{\infty}
sof doesn't intersect Y
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
−2x31−1x4=0- \frac{2}{x^{3} \sqrt{1 - \frac{1}{x^{4}}}} = 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⋅(3+2x4⋅(1−1x4))x41−1x4=0\frac{2 \cdot \left(3 + \frac{2}{x^{4} \cdot \left(1 - \frac{1}{x^{4}}\right)}\right)}{x^{4} \sqrt{1 - \frac{1}{x^{4}}}} = 0
Solve this equation
The roots of this equation
x1=−3343x_{1} = - \frac{3^{\frac{3}{4}}}{3}
x2=3343x_{2} = \frac{3^{\frac{3}{4}}}{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=0x_{1} = 0

lim⁡x→0−(2⋅(3+2x4⋅(1−1x4))x41−1x4)=−∞i\lim_{x \to 0^-}\left(\frac{2 \cdot \left(3 + \frac{2}{x^{4} \cdot \left(1 - \frac{1}{x^{4}}\right)}\right)}{x^{4} \sqrt{1 - \frac{1}{x^{4}}}}\right) = - \infty i
Let's take the limit
lim⁡x→0+(2⋅(3+2x4⋅(1−1x4))x41−1x4)=−∞i\lim_{x \to 0^+}\left(\frac{2 \cdot \left(3 + \frac{2}{x^{4} \cdot \left(1 - \frac{1}{x^{4}}\right)}\right)}{x^{4} \sqrt{1 - \frac{1}{x^{4}}}}\right) = - \infty i
Let's take the limit
- 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:
Have no bends at the whole real axis
Vertical asymptotes
Have:
x1=0x_{1} = 0
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞asin⁡(1⋅1x2)=0\lim_{x \to -\infty} \operatorname{asin}{\left(1 \cdot \frac{1}{x^{2}} \right)} = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞asin⁡(1⋅1x2)=0\lim_{x \to \infty} \operatorname{asin}{\left(1 \cdot \frac{1}{x^{2}} \right)} = 0
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=0y = 0
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of asin(1/x^2), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(asin⁡(1⋅1x2)x)=0\lim_{x \to -\infty}\left(\frac{\operatorname{asin}{\left(1 \cdot \frac{1}{x^{2}} \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(asin⁡(1⋅1x2)x)=0\lim_{x \to \infty}\left(\frac{\operatorname{asin}{\left(1 \cdot \frac{1}{x^{2}} \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⁡(1⋅1x2)=asin⁡(1⋅1x2)\operatorname{asin}{\left(1 \cdot \frac{1}{x^{2}} \right)} = \operatorname{asin}{\left(1 \cdot \frac{1}{x^{2}} \right)}
- Yes
asin⁡(1⋅1x2)=−asin⁡(1⋅1x2)\operatorname{asin}{\left(1 \cdot \frac{1}{x^{2}} \right)} = - \operatorname{asin}{\left(1 \cdot \frac{1}{x^{2}} \right)}
- No
so, the function
is
even
The graph
Graphing y = arcsin(1/x^2)
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