Mister Exam
Graphing y = x+ln(x)/x/x
The solution
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/log(x)\
|------|
\ x /
f(x) = x + --------
x
$$f{\left(x \right)} = x + \frac{\frac{1}{x} \log{\left(x \right)}}{x}$$
f = x + (log(x)/x)/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:
$$x + \frac{\frac{1}{x} \log{\left(x \right)}}{x} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = e^{- \frac{W\left(3\right)}{3}}$$
Numerical solution
$$x_{1} = 0.704709490254913$$
so we need to solve the equation:
$$x + \frac{\frac{1}{x} \log{\left(x \right)}}{x} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = e^{- \frac{W\left(3\right)}{3}}$$
Numerical solution
$$x_{1} = 0.704709490254913$$
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
$$1 + \frac{- \frac{\log{\left(x \right)}}{x^{2}} + \frac{1}{x^{2}}}{x} - \frac{\log{\left(x \right)}}{x^{3}} = 0$$
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
$$1 + \frac{- \frac{\log{\left(x \right)}}{x^{2}} + \frac{1}{x^{2}}}{x} - \frac{\log{\left(x \right)}}{x^{3}} = 0$$
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
$$\frac{6 \log{\left(x \right)} - 5}{x^{4}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = e^{\frac{5}{6}}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(\frac{6 \log{\left(x \right)} - 5}{x^{4}}\right) = -\infty$$
$$\lim_{x \to 0^+}\left(\frac{6 \log{\left(x \right)} - 5}{x^{4}}\right) = -\infty$$
- limits are equal, then skip the corresponding 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
$$\left[e^{\frac{5}{6}}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, e^{\frac{5}{6}}\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{6 \log{\left(x \right)} - 5}{x^{4}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = e^{\frac{5}{6}}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(\frac{6 \log{\left(x \right)} - 5}{x^{4}}\right) = -\infty$$
$$\lim_{x \to 0^+}\left(\frac{6 \log{\left(x \right)} - 5}{x^{4}}\right) = -\infty$$
- limits are equal, then skip the corresponding 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
$$\left[e^{\frac{5}{6}}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, e^{\frac{5}{6}}\right]$$
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
$$\lim_{x \to -\infty}\left(x + \frac{\frac{1}{x} \log{\left(x \right)}}{x}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(x + \frac{\frac{1}{x} \log{\left(x \right)}}{x}\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(x + \frac{\frac{1}{x} \log{\left(x \right)}}{x}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(x + \frac{\frac{1}{x} \log{\left(x \right)}}{x}\right) = \infty$$
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 + (log(x)/x)/x, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{x + \frac{\frac{1}{x} \log{\left(x \right)}}{x}}{x}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x$$
$$\lim_{x \to \infty}\left(\frac{x + \frac{\frac{1}{x} \log{\left(x \right)}}{x}}{x}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x$$
$$\lim_{x \to -\infty}\left(\frac{x + \frac{\frac{1}{x} \log{\left(x \right)}}{x}}{x}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x$$
$$\lim_{x \to \infty}\left(\frac{x + \frac{\frac{1}{x} \log{\left(x \right)}}{x}}{x}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x$$
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:
$$x + \frac{\frac{1}{x} \log{\left(x \right)}}{x} = - x + \frac{\log{\left(- x \right)}}{x^{2}}$$
- No
$$x + \frac{\frac{1}{x} \log{\left(x \right)}}{x} = x - \frac{\log{\left(- x \right)}}{x^{2}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$x + \frac{\frac{1}{x} \log{\left(x \right)}}{x} = - x + \frac{\log{\left(- x \right)}}{x^{2}}$$
- No
$$x + \frac{\frac{1}{x} \log{\left(x \right)}}{x} = x - \frac{\log{\left(- x \right)}}{x^{2}}$$
- No
so, the function
not is
neither even, nor odd