Mister Exam
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-
How to use it?
- Graphing y =:
- x^4+8x^3+16x^2
- -x^2-4x
- x^2-4x-2
- (x^2-3)/(x^2-1)
- Integral of d{x}:
-
sqrt(4-x^2)/x
- Derivative of:
-
sqrt(4-x^2)/x
- Limit of the function:
-
sqrt(4-x^2)/x
-
Identical expressions
- sqrt(four -x^ two)/x
- square root of (4 minus x squared ) divide by x
- square root of (four minus x to the power of two) divide by x
- √(4-x^2)/x
- sqrt(4-x2)/x
- sqrt4-x2/x
- sqrt(4-x²)/x
- sqrt(4-x to the power of 2)/x
- sqrt4-x^2/x
- sqrt(4-x^2) divide by x
-
Similar expressions
- sqrt(4+x^2)/x
Graphing y = sqrt(4-x^2)/x
The solution
You have entered
[src]
________
/ 2
\/ 4 - x
f(x) = -----------
x
$$f{\left(x \right)} = \frac{\sqrt{4 - x^{2}}}{x}$$
f = sqrt(4 - x^2)/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:
$$\frac{\sqrt{4 - x^{2}}}{x} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = -2$$
$$x_{2} = 2$$
Numerical solution
$$x_{1} = -2$$
$$x_{2} = 2$$
so we need to solve the equation:
$$\frac{\sqrt{4 - x^{2}}}{x} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = -2$$
$$x_{2} = 2$$
Numerical solution
$$x_{1} = -2$$
$$x_{2} = 2$$
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
$$- \frac{1}{\sqrt{4 - x^{2}}} - \frac{\sqrt{4 - x^{2}}}{x^{2}} = 0$$
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
$$- \frac{1}{\sqrt{4 - x^{2}}} - \frac{\sqrt{4 - x^{2}}}{x^{2}} = 0$$
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
$$\frac{\frac{\frac{x^{2}}{x^{2} - 4} - 1}{\sqrt{4 - x^{2}}} + \frac{2}{\sqrt{4 - x^{2}}} + \frac{2 \sqrt{4 - x^{2}}}{x^{2}}}{x} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{2 \sqrt{6}}{3}$$
$$x_{2} = \frac{2 \sqrt{6}}{3}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(\frac{\frac{\frac{x^{2}}{x^{2} - 4} - 1}{\sqrt{4 - x^{2}}} + \frac{2}{\sqrt{4 - x^{2}}} + \frac{2 \sqrt{4 - x^{2}}}{x^{2}}}{x}\right) = -\infty$$
$$\lim_{x \to 0^+}\left(\frac{\frac{\frac{x^{2}}{x^{2} - 4} - 1}{\sqrt{4 - x^{2}}} + \frac{2}{\sqrt{4 - x^{2}}} + \frac{2 \sqrt{4 - x^{2}}}{x^{2}}}{x}\right) = \infty$$
- 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:
Concave at the intervals
$$\left(-\infty, - \frac{2 \sqrt{6}}{3}\right]$$
Convex at the intervals
$$\left[\frac{2 \sqrt{6}}{3}, \infty\right)$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{\frac{\frac{x^{2}}{x^{2} - 4} - 1}{\sqrt{4 - x^{2}}} + \frac{2}{\sqrt{4 - x^{2}}} + \frac{2 \sqrt{4 - x^{2}}}{x^{2}}}{x} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{2 \sqrt{6}}{3}$$
$$x_{2} = \frac{2 \sqrt{6}}{3}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(\frac{\frac{\frac{x^{2}}{x^{2} - 4} - 1}{\sqrt{4 - x^{2}}} + \frac{2}{\sqrt{4 - x^{2}}} + \frac{2 \sqrt{4 - x^{2}}}{x^{2}}}{x}\right) = -\infty$$
$$\lim_{x \to 0^+}\left(\frac{\frac{\frac{x^{2}}{x^{2} - 4} - 1}{\sqrt{4 - x^{2}}} + \frac{2}{\sqrt{4 - x^{2}}} + \frac{2 \sqrt{4 - x^{2}}}{x^{2}}}{x}\right) = \infty$$
- 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:
Concave at the intervals
$$\left(-\infty, - \frac{2 \sqrt{6}}{3}\right]$$
Convex at the intervals
$$\left[\frac{2 \sqrt{6}}{3}, \infty\right)$$
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
$$\lim_{x \to -\infty}\left(\frac{\sqrt{4 - x^{2}}}{x}\right) = - i$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = - i$$
$$\lim_{x \to \infty}\left(\frac{\sqrt{4 - x^{2}}}{x}\right) = i$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = i$$
$$\lim_{x \to -\infty}\left(\frac{\sqrt{4 - x^{2}}}{x}\right) = - i$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = - i$$
$$\lim_{x \to \infty}\left(\frac{\sqrt{4 - x^{2}}}{x}\right) = i$$
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(4 - x^2)/x, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\sqrt{4 - x^{2}}}{x^{2}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sqrt{4 - x^{2}}}{x^{2}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\sqrt{4 - x^{2}}}{x^{2}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sqrt{4 - x^{2}}}{x^{2}}\right) = 0$$
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:
$$\frac{\sqrt{4 - x^{2}}}{x} = - \frac{\sqrt{4 - x^{2}}}{x}$$
- No
$$\frac{\sqrt{4 - x^{2}}}{x} = \frac{\sqrt{4 - x^{2}}}{x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{\sqrt{4 - x^{2}}}{x} = - \frac{\sqrt{4 - x^{2}}}{x}$$
- No
$$\frac{\sqrt{4 - x^{2}}}{x} = \frac{\sqrt{4 - x^{2}}}{x}$$
- No
so, the function
not is
neither even, nor odd