Mister Exam
Graphing y = x/(sqrt(x-2))^(2/3)
The solution
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x
f(x) = ------------
2/3
_______
\/ x - 2
$$f{\left(x \right)} = \frac{x}{\left(\sqrt{x - 2}\right)^{\frac{2}{3}}}$$
f = x/(sqrt(x - 2))^(2/3)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 2$$
$$x_{1} = 2$$
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:
$$\frac{x}{\left(\sqrt{x - 2}\right)^{\frac{2}{3}}} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 0$$
Numerical solution
$$x_{1} = 0$$
so we need to solve the equation:
$$\frac{x}{\left(\sqrt{x - 2}\right)^{\frac{2}{3}}} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 0$$
Numerical solution
$$x_{1} = 0$$
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
$$\frac{2 \left(\frac{2 x}{x - 2} - 3\right)}{9 \left(x - 2\right)^{\frac{4}{3}}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 6$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 2$$
$$\lim_{x \to 2^-}\left(\frac{2 \left(\frac{2 x}{x - 2} - 3\right)}{9 \left(x - 2\right)^{\frac{4}{3}}}\right) = - \infty \left(-1\right)^{\frac{2}{3}}$$
$$\lim_{x \to 2^+}\left(\frac{2 \left(\frac{2 x}{x - 2} - 3\right)}{9 \left(x - 2\right)^{\frac{4}{3}}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 2$$
- 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
$$\left(-\infty, 6\right]$$
Convex at the intervals
$$\left[6, \infty\right)$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{2 \left(\frac{2 x}{x - 2} - 3\right)}{9 \left(x - 2\right)^{\frac{4}{3}}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 6$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 2$$
$$\lim_{x \to 2^-}\left(\frac{2 \left(\frac{2 x}{x - 2} - 3\right)}{9 \left(x - 2\right)^{\frac{4}{3}}}\right) = - \infty \left(-1\right)^{\frac{2}{3}}$$
$$\lim_{x \to 2^+}\left(\frac{2 \left(\frac{2 x}{x - 2} - 3\right)}{9 \left(x - 2\right)^{\frac{4}{3}}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 2$$
- 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
$$\left(-\infty, 6\right]$$
Convex at the intervals
$$\left[6, \infty\right)$$
Vertical asymptotes
Have:
$$x_{1} = 2$$
$$x_{1} = 2$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{x}{\left(\sqrt{x - 2}\right)^{\frac{2}{3}}}\right) = \infty \left(-1\right)^{\frac{2}{3}}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \infty \left(-1\right)^{\frac{2}{3}}$$
$$\lim_{x \to \infty}\left(\frac{x}{\left(\sqrt{x - 2}\right)^{\frac{2}{3}}}\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\frac{x}{\left(\sqrt{x - 2}\right)^{\frac{2}{3}}}\right) = \infty \left(-1\right)^{\frac{2}{3}}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \infty \left(-1\right)^{\frac{2}{3}}$$
$$\lim_{x \to \infty}\left(\frac{x}{\left(\sqrt{x - 2}\right)^{\frac{2}{3}}}\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of x/(sqrt(x - 2))^(2/3), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty} \frac{1}{\sqrt[3]{x - 2}} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty} \frac{1}{\sqrt[3]{x - 2}} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty} \frac{1}{\sqrt[3]{x - 2}} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty} \frac{1}{\sqrt[3]{x - 2}} = 0$$
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:
$$\frac{x}{\left(\sqrt{x - 2}\right)^{\frac{2}{3}}} = - \frac{x}{\sqrt[3]{- x - 2}}$$
- No
$$\frac{x}{\left(\sqrt{x - 2}\right)^{\frac{2}{3}}} = \frac{x}{\sqrt[3]{- x - 2}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{x}{\left(\sqrt{x - 2}\right)^{\frac{2}{3}}} = - \frac{x}{\sqrt[3]{- x - 2}}$$
- No
$$\frac{x}{\left(\sqrt{x - 2}\right)^{\frac{2}{3}}} = \frac{x}{\sqrt[3]{- x - 2}}$$
- No
so, the function
not is
neither even, nor odd