Mister Exam
Other calculators
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- The intersection points of functions
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How to use it?
- Graphing y =:
- x^3-((x^4)/4)
- x^3+6*x^2+9*x+4
- -x^3+3x^2+9x-12
- x^3+4x
- Integral of d{x}:
-
sqrt(x-1)/sqrt(x+1)
-
Identical expressions
- sqrt(x- one)/sqrt(x+ one)
- square root of (x minus 1) divide by square root of (x plus 1)
- square root of (x minus one) divide by square root of (x plus one)
- √(x-1)/√(x+1)
- sqrtx-1/sqrtx+1
- sqrt(x-1) divide by sqrt(x+1)
-
Similar expressions
- sqrt(x)-1/sqrt(x)+1/x
- sqrt(x+1)/sqrt(x+1)
- sqrt(x-1)/sqrt(x-1)
Graphing y = sqrt(x-1)/sqrt(x+1)
The solution
You have entered
[src]
_______
\/ x - 1
f(x) = ---------
_______
\/ x + 1
$$f{\left(x \right)} = \frac{\sqrt{x - 1}}{\sqrt{x + 1}}$$
f = sqrt(x - 1)/sqrt(x + 1)
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:
$$\frac{\sqrt{x - 1}}{\sqrt{x + 1}} = 0$$
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:
$$\frac{\sqrt{x - 1}}{\sqrt{x + 1}} = 0$$
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
$$- \frac{\sqrt{x - 1}}{2 \left(x + 1\right)^{\frac{3}{2}}} + \frac{1}{2 \sqrt{x - 1} \sqrt{x + 1}} = 0$$
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
$$- \frac{\sqrt{x - 1}}{2 \left(x + 1\right)^{\frac{3}{2}}} + \frac{1}{2 \sqrt{x - 1} \sqrt{x + 1}} = 0$$
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
$$\frac{\frac{3 \sqrt{x - 1}}{\left(x + 1\right)^{2}} - \frac{2}{\sqrt{x - 1} \left(x + 1\right)} - \frac{1}{\left(x - 1\right)^{\frac{3}{2}}}}{4 \sqrt{x + 1}} = 0$$
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:
$$x_{1} = -1$$
$$\lim_{x \to -1^-}\left(\frac{\frac{3 \sqrt{x - 1}}{\left(x + 1\right)^{2}} - \frac{2}{\sqrt{x - 1} \left(x + 1\right)} - \frac{1}{\left(x - 1\right)^{\frac{3}{2}}}}{4 \sqrt{x + 1}}\right) = \infty$$
$$\lim_{x \to -1^+}\left(\frac{\frac{3 \sqrt{x - 1}}{\left(x + 1\right)^{2}} - \frac{2}{\sqrt{x - 1} \left(x + 1\right)} - \frac{1}{\left(x - 1\right)^{\frac{3}{2}}}}{4 \sqrt{x + 1}}\right) = \infty i$$
- 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:
Have no bends at the whole real axis
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{\frac{3 \sqrt{x - 1}}{\left(x + 1\right)^{2}} - \frac{2}{\sqrt{x - 1} \left(x + 1\right)} - \frac{1}{\left(x - 1\right)^{\frac{3}{2}}}}{4 \sqrt{x + 1}} = 0$$
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:
$$x_{1} = -1$$
$$\lim_{x \to -1^-}\left(\frac{\frac{3 \sqrt{x - 1}}{\left(x + 1\right)^{2}} - \frac{2}{\sqrt{x - 1} \left(x + 1\right)} - \frac{1}{\left(x - 1\right)^{\frac{3}{2}}}}{4 \sqrt{x + 1}}\right) = \infty$$
$$\lim_{x \to -1^+}\left(\frac{\frac{3 \sqrt{x - 1}}{\left(x + 1\right)^{2}} - \frac{2}{\sqrt{x - 1} \left(x + 1\right)} - \frac{1}{\left(x - 1\right)^{\frac{3}{2}}}}{4 \sqrt{x + 1}}\right) = \infty i$$
- 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:
Have no bends at the whole real axis
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
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x - 1}}{\sqrt{x + 1}}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{\sqrt{x - 1}}{\sqrt{x + 1}}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x - 1}}{\sqrt{x + 1}}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{\sqrt{x - 1}}{\sqrt{x + 1}}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of sqrt(x - 1)/sqrt(x + 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x - 1}}{x \sqrt{x + 1}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sqrt{x - 1}}{x \sqrt{x + 1}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x - 1}}{x \sqrt{x + 1}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sqrt{x - 1}}{x \sqrt{x + 1}}\right) = 0$$
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:
$$\frac{\sqrt{x - 1}}{\sqrt{x + 1}} = \frac{\sqrt{- x - 1}}{\sqrt{1 - x}}$$
- No
$$\frac{\sqrt{x - 1}}{\sqrt{x + 1}} = - \frac{\sqrt{- x - 1}}{\sqrt{1 - x}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{\sqrt{x - 1}}{\sqrt{x + 1}} = \frac{\sqrt{- x - 1}}{\sqrt{1 - x}}$$
- No
$$\frac{\sqrt{x - 1}}{\sqrt{x + 1}} = - \frac{\sqrt{- x - 1}}{\sqrt{1 - x}}$$
- No
so, the function
not is
neither even, nor odd