Mister Exam
Graphing y = sqrt(16-x^2)/x-2
The solution
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/ 2
\/ 16 - x
f(x) = ------------ - 2
x
$$f{\left(x \right)} = -2 + \frac{\sqrt{16 - x^{2}}}{x}$$
f = -2 + sqrt(16 - 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:
$$-2 + \frac{\sqrt{16 - x^{2}}}{x} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = \frac{4 \sqrt{5}}{5}$$
Numerical solution
$$x_{1} = 1.78885438199983$$
so we need to solve the equation:
$$-2 + \frac{\sqrt{16 - x^{2}}}{x} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = \frac{4 \sqrt{5}}{5}$$
Numerical solution
$$x_{1} = 1.78885438199983$$
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
$$- \frac{1}{\sqrt{16 - x^{2}}} - \frac{\sqrt{16 - x^{2}}}{x^{2}} = 0$$
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
$$- \frac{1}{\sqrt{16 - x^{2}}} - \frac{\sqrt{16 - x^{2}}}{x^{2}} = 0$$
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
$$- \frac{x}{\left(16 - x^{2}\right)^{\frac{3}{2}}} + \frac{1}{x \sqrt{16 - x^{2}}} + \frac{2 \sqrt{16 - x^{2}}}{x^{3}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{4 \sqrt{6}}{3}$$
$$x_{2} = \frac{4 \sqrt{6}}{3}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(- \frac{x}{\left(16 - x^{2}\right)^{\frac{3}{2}}} + \frac{1}{x \sqrt{16 - x^{2}}} + \frac{2 \sqrt{16 - x^{2}}}{x^{3}}\right) = -\infty$$
$$\lim_{x \to 0^+}\left(- \frac{x}{\left(16 - x^{2}\right)^{\frac{3}{2}}} + \frac{1}{x \sqrt{16 - x^{2}}} + \frac{2 \sqrt{16 - x^{2}}}{x^{3}}\right) = \infty$$
- 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
$$\left(-\infty, - \frac{4 \sqrt{6}}{3}\right]$$
Convex at the intervals
$$\left[\frac{4 \sqrt{6}}{3}, \infty\right)$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$- \frac{x}{\left(16 - x^{2}\right)^{\frac{3}{2}}} + \frac{1}{x \sqrt{16 - x^{2}}} + \frac{2 \sqrt{16 - x^{2}}}{x^{3}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{4 \sqrt{6}}{3}$$
$$x_{2} = \frac{4 \sqrt{6}}{3}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(- \frac{x}{\left(16 - x^{2}\right)^{\frac{3}{2}}} + \frac{1}{x \sqrt{16 - x^{2}}} + \frac{2 \sqrt{16 - x^{2}}}{x^{3}}\right) = -\infty$$
$$\lim_{x \to 0^+}\left(- \frac{x}{\left(16 - x^{2}\right)^{\frac{3}{2}}} + \frac{1}{x \sqrt{16 - x^{2}}} + \frac{2 \sqrt{16 - x^{2}}}{x^{3}}\right) = \infty$$
- 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
$$\left(-\infty, - \frac{4 \sqrt{6}}{3}\right]$$
Convex at the intervals
$$\left[\frac{4 \sqrt{6}}{3}, \infty\right)$$
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
$$\lim_{x \to -\infty}\left(-2 + \frac{\sqrt{16 - x^{2}}}{x}\right) = -2 - i$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = -2 - i$$
$$\lim_{x \to \infty}\left(-2 + \frac{\sqrt{16 - x^{2}}}{x}\right) = -2 + i$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = -2 + i$$
$$\lim_{x \to -\infty}\left(-2 + \frac{\sqrt{16 - x^{2}}}{x}\right) = -2 - i$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = -2 - i$$
$$\lim_{x \to \infty}\left(-2 + \frac{\sqrt{16 - x^{2}}}{x}\right) = -2 + i$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = -2 + i$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of sqrt(16 - x^2)/x - 2, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{-2 + \frac{\sqrt{16 - x^{2}}}{x}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{-2 + \frac{\sqrt{16 - x^{2}}}{x}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{-2 + \frac{\sqrt{16 - x^{2}}}{x}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{-2 + \frac{\sqrt{16 - x^{2}}}{x}}{x}\right) = 0$$
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:
$$-2 + \frac{\sqrt{16 - x^{2}}}{x} = -2 - \frac{\sqrt{16 - x^{2}}}{x}$$
- No
$$-2 + \frac{\sqrt{16 - x^{2}}}{x} = 2 + \frac{\sqrt{16 - x^{2}}}{x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$-2 + \frac{\sqrt{16 - x^{2}}}{x} = -2 - \frac{\sqrt{16 - x^{2}}}{x}$$
- No
$$-2 + \frac{\sqrt{16 - x^{2}}}{x} = 2 + \frac{\sqrt{16 - x^{2}}}{x}$$
- No
so, the function
not is
neither even, nor odd