Mister Exam
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-
How to use it?
- Graphing y =:
- x^3-x^2-40x+3
-
log2x
- 3x^2+5x-6
- 3x^2-2
- Limit of the function:
-
sqrt((1+x)/x)
-
Identical expressions
- sqrt((one +x)/x)
- square root of ((1 plus x) divide by x)
- square root of ((one plus x) divide by x)
- √((1+x)/x)
- sqrt1+x/x
- sqrt((1+x) divide by x)
-
Similar expressions
- sqrt((1-x)/x)
Graphing y = sqrt((1+x)/x)
The solution
You have entered
[src]
_______
/ 1 + x
f(x) = / -----
\/ x
$$f{\left(x \right)} = \sqrt{\frac{x + 1}{x}}$$
f = sqrt((x + 1)/x)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 0$$
$$x_{1} = 0$$
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:
$$\sqrt{\frac{x + 1}{x}} = 0$$
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:
$$\sqrt{\frac{x + 1}{x}} = 0$$
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
$$\frac{x \sqrt{\frac{x + 1}{x}} \left(\frac{1}{2 x} - \frac{x + 1}{2 x^{2}}\right)}{x + 1} = 0$$
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
$$\frac{x \sqrt{\frac{x + 1}{x}} \left(\frac{1}{2 x} - \frac{x + 1}{2 x^{2}}\right)}{x + 1} = 0$$
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
$$\frac{\sqrt{\frac{x + 1}{x}} \left(1 - \frac{x + 1}{x}\right) \left(\frac{1 - \frac{x + 1}{x}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x}\right)}{4 \left(x + 1\right)} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{3}{4}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(\frac{\sqrt{\frac{x + 1}{x}} \left(1 - \frac{x + 1}{x}\right) \left(\frac{1 - \frac{x + 1}{x}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x}\right)}{4 \left(x + 1\right)}\right) = \infty i$$
$$\lim_{x \to 0^+}\left(\frac{\sqrt{\frac{x + 1}{x}} \left(1 - \frac{x + 1}{x}\right) \left(\frac{1 - \frac{x + 1}{x}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x}\right)}{4 \left(x + 1\right)}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 0$$
- 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
$$\frac{\sqrt{\frac{x + 1}{x}} \left(1 - \frac{x + 1}{x}\right) \left(\frac{1 - \frac{x + 1}{x}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x}\right)}{4 \left(x + 1\right)} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{3}{4}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(\frac{\sqrt{\frac{x + 1}{x}} \left(1 - \frac{x + 1}{x}\right) \left(\frac{1 - \frac{x + 1}{x}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x}\right)}{4 \left(x + 1\right)}\right) = \infty i$$
$$\lim_{x \to 0^+}\left(\frac{\sqrt{\frac{x + 1}{x}} \left(1 - \frac{x + 1}{x}\right) \left(\frac{1 - \frac{x + 1}{x}}{x + 1} - \frac{2}{x + 1} - \frac{2}{x}\right)}{4 \left(x + 1\right)}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 0$$
- 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:
$$x_{1} = 0$$
$$x_{1} = 0$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \sqrt{\frac{x + 1}{x}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty} \sqrt{\frac{x + 1}{x}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
$$\lim_{x \to -\infty} \sqrt{\frac{x + 1}{x}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty} \sqrt{\frac{x + 1}{x}} = 1$$
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((1 + x)/x), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\sqrt{\frac{x + 1}{x}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sqrt{\frac{x + 1}{x}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\sqrt{\frac{x + 1}{x}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sqrt{\frac{x + 1}{x}}}{x}\right) = 0$$
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:
$$\sqrt{\frac{x + 1}{x}} = \sqrt{- \frac{1 - x}{x}}$$
- No
$$\sqrt{\frac{x + 1}{x}} = - \sqrt{- \frac{1 - x}{x}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\sqrt{\frac{x + 1}{x}} = \sqrt{- \frac{1 - x}{x}}$$
- No
$$\sqrt{\frac{x + 1}{x}} = - \sqrt{- \frac{1 - x}{x}}$$
- No
so, the function
not is
neither even, nor odd