Mister Exam
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-
How to use it?
- Graphing y =:
- x+
- x^5-5x^4+5x^3+1
- x^4+8x^2+9
- x^4-2x^2+8
- Derivative of:
-
sqrt((1+x)/(1-x))
- Limit of the function:
-
sqrt((1+x)/(1-x))
- Integral of d{x}:
- sqrt((1+x)/(1-x))
-
Identical expressions
- sqrt((one +x)/(one -x))
- square root of ((1 plus x) divide by (1 minus x))
- square root of ((one plus x) divide by (one minus x))
- √((1+x)/(1-x))
- sqrt1+x/1-x
- sqrt((1+x) divide by (1-x))
-
Similar expressions
- sqrt((1-x)/(1-x))
- sqrt((1+x)/(1+x))
Graphing y = sqrt((1+x)/(1-x))
The solution
You have entered
[src]
_______
/ 1 + x
f(x) = / -----
\/ 1 - x
$$f{\left(x \right)} = \sqrt{\frac{x + 1}{1 - x}}$$
f = sqrt((x + 1)/(1 - x))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 1$$
$$x_{1} = 1$$
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:
$$\sqrt{\frac{x + 1}{1 - x}} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = -1$$
Numerical solution
$$x_{1} = -1$$
so we need to solve the equation:
$$\sqrt{\frac{x + 1}{1 - x}} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = -1$$
Numerical solution
$$x_{1} = -1$$
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
$$\frac{\sqrt{\frac{x + 1}{1 - x}} \left(1 - x\right) \left(\frac{1}{2 \left(1 - x\right)} + \frac{x + 1}{2 \left(1 - x\right)^{2}}\right)}{x + 1} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(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
$$\frac{\sqrt{\frac{x + 1}{1 - x}} \left(1 - x\right) \left(\frac{1}{2 \left(1 - x\right)} + \frac{x + 1}{2 \left(1 - x\right)^{2}}\right)}{x + 1} = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{\sqrt{- \frac{x + 1}{x - 1}} \left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1 - \frac{x + 1}{x - 1}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x - 1}\right)}{4 \left(x + 1\right)} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{1}{2}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 1$$
$$\lim_{x \to 1^-}\left(\frac{\sqrt{- \frac{x + 1}{x - 1}} \left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1 - \frac{x + 1}{x - 1}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x - 1}\right)}{4 \left(x + 1\right)}\right) = \infty$$
$$\lim_{x \to 1^+}\left(\frac{\sqrt{- \frac{x + 1}{x - 1}} \left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1 - \frac{x + 1}{x - 1}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x - 1}\right)}{4 \left(x + 1\right)}\right) = \infty i$$
- the limits are not equal, so
$$x_{1} = 1$$
- 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
$$\left[- \frac{1}{2}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, - \frac{1}{2}\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{\sqrt{- \frac{x + 1}{x - 1}} \left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1 - \frac{x + 1}{x - 1}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x - 1}\right)}{4 \left(x + 1\right)} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{1}{2}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 1$$
$$\lim_{x \to 1^-}\left(\frac{\sqrt{- \frac{x + 1}{x - 1}} \left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1 - \frac{x + 1}{x - 1}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x - 1}\right)}{4 \left(x + 1\right)}\right) = \infty$$
$$\lim_{x \to 1^+}\left(\frac{\sqrt{- \frac{x + 1}{x - 1}} \left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1 - \frac{x + 1}{x - 1}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x - 1}\right)}{4 \left(x + 1\right)}\right) = \infty i$$
- the limits are not equal, so
$$x_{1} = 1$$
- 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
$$\left[- \frac{1}{2}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, - \frac{1}{2}\right]$$
Vertical asymptotes
Have:
$$x_{1} = 1$$
$$x_{1} = 1$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \sqrt{\frac{x + 1}{1 - x}} = i$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = i$$
$$\lim_{x \to \infty} \sqrt{\frac{x + 1}{1 - x}} = i$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = i$$
$$\lim_{x \to -\infty} \sqrt{\frac{x + 1}{1 - x}} = i$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = i$$
$$\lim_{x \to \infty} \sqrt{\frac{x + 1}{1 - x}} = i$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = i$$
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
$$\lim_{x \to -\infty}\left(\frac{\sqrt{\frac{x + 1}{1 - x}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sqrt{\frac{x + 1}{1 - x}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\sqrt{\frac{x + 1}{1 - x}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sqrt{\frac{x + 1}{1 - x}}}{x}\right) = 0$$
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:
$$\sqrt{\frac{x + 1}{1 - x}} = \sqrt{\frac{1 - x}{x + 1}}$$
- No
$$\sqrt{\frac{x + 1}{1 - x}} = - \sqrt{\frac{1 - x}{x + 1}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\sqrt{\frac{x + 1}{1 - x}} = \sqrt{\frac{1 - x}{x + 1}}$$
- No
$$\sqrt{\frac{x + 1}{1 - x}} = - \sqrt{\frac{1 - x}{x + 1}}$$
- No
so, the function
not is
neither even, nor odd