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Plotting a function graph/ sqrt((1+x)/(1-x))

Graphing y = sqrt((1+x)/(1-x))

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
           _______
          / 1 + x 
f(x) =   /  ----- 
       \/   1 - x
f(x)=x+11−xf{\left(x \right)} = \sqrt{\frac{x + 1}{1 - x}} 
f = sqrt((x + 1)/(1 - x))
The graph of the function
02468-8-6-4-2-1010010
The domain of the function
The points at which the function is not precisely defined:
x1=1x_{1} = 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:
x+11−x=0\sqrt{\frac{x + 1}{1 - x}} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−1x_{1} = -1
Numerical solution
x1=−1x_{1} = -1
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to sqrt((1 + x)/(1 - x)).
11−0\sqrt{\frac{1}{1 - 0}}
The result:
f(0)=1f{\left(0 \right)} = 1
The point:
(0, 1)
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
x+11−x(1−x)(12(1−x)+x+12(1−x)2)x+1=0\frac{\sqrt{\frac{x + 1}{1 - x}} \left(1 - x\right) \left(\frac{1}{2 \left(1 - x\right)} + \frac{x + 1}{2 \left(1 - x\right)^{2}}\right)}{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
−x+1x−1(1−x+1x−1)(1−x+1x−1x+1−2x+1−2x−1)4(x+1)=0\frac{\sqrt{- \frac{x + 1}{x - 1}} \left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1 - \frac{x + 1}{x - 1}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x - 1}\right)}{4 \left(x + 1\right)} = 0
Solve this equation
The roots of this equation
x1=−12x_{1} = - \frac{1}{2}
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

lim⁡x→1−(−x+1x−1(1−x+1x−1)(1−x+1x−1x+1−2x+1−2x−1)4(x+1))=∞\lim_{x \to 1^-}\left(\frac{\sqrt{- \frac{x + 1}{x - 1}} \left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1 - \frac{x + 1}{x - 1}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x - 1}\right)}{4 \left(x + 1\right)}\right) = \infty
lim⁡x→1+(−x+1x−1(1−x+1x−1)(1−x+1x−1x+1−2x+1−2x−1)4(x+1))=∞i\lim_{x \to 1^+}\left(\frac{\sqrt{- \frac{x + 1}{x - 1}} \left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1 - \frac{x + 1}{x - 1}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x - 1}\right)}{4 \left(x + 1\right)}\right) = \infty i
- the limits are not equal, so
x1=1x_{1} = 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:
Concave at the intervals
[−12,∞)\left[- \frac{1}{2}, \infty\right)
Convex at the intervals
(−∞,−12]\left(-\infty, - \frac{1}{2}\right]
Vertical asymptotes
Have:
x1=1x_{1} = 1
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞x+11−x=i\lim_{x \to -\infty} \sqrt{\frac{x + 1}{1 - x}} = i
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=iy = i
lim⁡x→∞x+11−x=i\lim_{x \to \infty} \sqrt{\frac{x + 1}{1 - x}} = i
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=iy = i
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of sqrt((1 + x)/(1 - x)), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(x+11−xx)=0\lim_{x \to -\infty}\left(\frac{\sqrt{\frac{x + 1}{1 - x}}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(x+11−xx)=0\lim_{x \to \infty}\left(\frac{\sqrt{\frac{x + 1}{1 - x}}}{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:
x+11−x=1−xx+1\sqrt{\frac{x + 1}{1 - x}} = \sqrt{\frac{1 - x}{x + 1}}
- No
x+11−x=−1−xx+1\sqrt{\frac{x + 1}{1 - x}} = - \sqrt{\frac{1 - x}{x + 1}}
- No
so, the function
not is
neither even, nor odd
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