Mister Exam

Other calculators

  • How to use it?

  • Graphing y =:
  • 5x^2-3x-1
  • -3x+5
  • 3/2x^2-x^3
  • 3x^2-12x

  • Identical expressions

  • xarccos(x/(one -x^ two))
  • xarc co sinus of e of (x divide by (1 minus x squared ))
  • xarc co sinus of e of (x divide by (one minus x to the power of two))
  • xarccos(x/(1-x2))
  • xarccosx/1-x2
  • xarccos(x/(1-x²))
  • xarccos(x/(1-x to the power of 2))
  • xarccosx/1-x^2
  • xarccos(x divide by (1-x^2))

  • Similar expressions

  • xarccos(x/(1+x^2))
Plotting a function graph/ xarccos(x/(1-x^2))

Graphing y = xarccos(x/(1-x^2))

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
             /  x   \
f(x) = x*acos|------|
             |     2|
             \1 - x /
f(x)=xacos⁡(x1−x2)f{\left(x \right)} = x \operatorname{acos}{\left(\frac{x}{1 - x^{2}} \right)} 
f = x*acos(x/(1 - x^2))
The graph of the function
02468-8-6-4-2-1010-5050
The domain of the function
The points at which the function is not precisely defined:
x1=−1x_{1} = -1
x2=1x_{2} = 1
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:
xacos⁡(x1−x2)=0x \operatorname{acos}{\left(\frac{x}{1 - x^{2}} \right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=0x_{1} = 0
x2=−12+52x_{2} = - \frac{1}{2} + \frac{\sqrt{5}}{2}
x3=−52−12x_{3} = - \frac{\sqrt{5}}{2} - \frac{1}{2}
Numerical solution
x1=0x_{1} = 0
x2=0.618033988749895x_{2} = 0.618033988749895
x3=−1.61803398874989x_{3} = -1.61803398874989
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to x*acos(x/(1 - x^2)).
0acos⁡(01−02)0 \operatorname{acos}{\left(\frac{0}{1 - 0^{2}} \right)}
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
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
−x2(−8x2x2−1+6+(2x2x2−1−1)2(x2−1)(x2(x2−1)2−1))x2−1+4x2x2−1−2(x2−1)−x2(1−x2)2+1=0- \frac{\frac{x^{2} \left(- \frac{8 x^{2}}{x^{2} - 1} + 6 + \frac{\left(\frac{2 x^{2}}{x^{2} - 1} - 1\right)^{2}}{\left(x^{2} - 1\right) \left(\frac{x^{2}}{\left(x^{2} - 1\right)^{2}} - 1\right)}\right)}{x^{2} - 1} + \frac{4 x^{2}}{x^{2} - 1} - 2}{\left(x^{2} - 1\right) \sqrt{- \frac{x^{2}}{\left(1 - x^{2}\right)^{2}} + 1}} = 0
Solve this equation
The roots of this equation
x1=−1.35560118154383x_{1} = -1.35560118154383
x2=−0.69551980859689x_{2} = -0.69551980859689
x3=0.69551980859689x_{3} = 0.69551980859689
x4=1.35560118154383x_{4} = 1.35560118154383
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=−1x_{1} = -1
x2=1x_{2} = 1

lim⁡x→−1−(−x2(−8x2x2−1+6+(2x2x2−1−1)2(x2−1)(x2(x2−1)2−1))x2−1+4x2x2−1−2(x2−1)−x2(1−x2)2+1)=−∞i\lim_{x \to -1^-}\left(- \frac{\frac{x^{2} \left(- \frac{8 x^{2}}{x^{2} - 1} + 6 + \frac{\left(\frac{2 x^{2}}{x^{2} - 1} - 1\right)^{2}}{\left(x^{2} - 1\right) \left(\frac{x^{2}}{\left(x^{2} - 1\right)^{2}} - 1\right)}\right)}{x^{2} - 1} + \frac{4 x^{2}}{x^{2} - 1} - 2}{\left(x^{2} - 1\right) \sqrt{- \frac{x^{2}}{\left(1 - x^{2}\right)^{2}} + 1}}\right) = - \infty i
lim⁡x→−1+(−x2(−8x2x2−1+6+(2x2x2−1−1)2(x2−1)(x2(x2−1)2−1))x2−1+4x2x2−1−2(x2−1)−x2(1−x2)2+1)=∞i\lim_{x \to -1^+}\left(- \frac{\frac{x^{2} \left(- \frac{8 x^{2}}{x^{2} - 1} + 6 + \frac{\left(\frac{2 x^{2}}{x^{2} - 1} - 1\right)^{2}}{\left(x^{2} - 1\right) \left(\frac{x^{2}}{\left(x^{2} - 1\right)^{2}} - 1\right)}\right)}{x^{2} - 1} + \frac{4 x^{2}}{x^{2} - 1} - 2}{\left(x^{2} - 1\right) \sqrt{- \frac{x^{2}}{\left(1 - x^{2}\right)^{2}} + 1}}\right) = \infty i
- the limits are not equal, so
x1=−1x_{1} = -1
- is an inflection point
lim⁡x→1−(−x2(−8x2x2−1+6+(2x2x2−1−1)2(x2−1)(x2(x2−1)2−1))x2−1+4x2x2−1−2(x2−1)−x2(1−x2)2+1)=∞i\lim_{x \to 1^-}\left(- \frac{\frac{x^{2} \left(- \frac{8 x^{2}}{x^{2} - 1} + 6 + \frac{\left(\frac{2 x^{2}}{x^{2} - 1} - 1\right)^{2}}{\left(x^{2} - 1\right) \left(\frac{x^{2}}{\left(x^{2} - 1\right)^{2}} - 1\right)}\right)}{x^{2} - 1} + \frac{4 x^{2}}{x^{2} - 1} - 2}{\left(x^{2} - 1\right) \sqrt{- \frac{x^{2}}{\left(1 - x^{2}\right)^{2}} + 1}}\right) = \infty i
lim⁡x→1+(−x2(−8x2x2−1+6+(2x2x2−1−1)2(x2−1)(x2(x2−1)2−1))x2−1+4x2x2−1−2(x2−1)−x2(1−x2)2+1)=−∞i\lim_{x \to 1^+}\left(- \frac{\frac{x^{2} \left(- \frac{8 x^{2}}{x^{2} - 1} + 6 + \frac{\left(\frac{2 x^{2}}{x^{2} - 1} - 1\right)^{2}}{\left(x^{2} - 1\right) \left(\frac{x^{2}}{\left(x^{2} - 1\right)^{2}} - 1\right)}\right)}{x^{2} - 1} + \frac{4 x^{2}}{x^{2} - 1} - 2}{\left(x^{2} - 1\right) \sqrt{- \frac{x^{2}}{\left(1 - x^{2}\right)^{2}} + 1}}\right) = - \infty i
- the limits are not equal, so
x2=1x_{2} = 1
- 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=−1x_{1} = -1
x2=1x_{2} = 1
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:
xacos⁡(x1−x2)=−xacos⁡(−x1−x2)x \operatorname{acos}{\left(\frac{x}{1 - x^{2}} \right)} = - x \operatorname{acos}{\left(- \frac{x}{1 - x^{2}} \right)}
- No
xacos⁡(x1−x2)=xacos⁡(−x1−x2)x \operatorname{acos}{\left(\frac{x}{1 - x^{2}} \right)} = x \operatorname{acos}{\left(- \frac{x}{1 - x^{2}} \right)}
- No
so, the function
not is
neither even, nor odd
    © Mister Exam – Calculator