Mister Exam

Graphing y = x+(lnx)/x

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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           log(x)
f(x) = x + ------
             x   
f(x)=x+log(x)xf{\left(x \right)} = x + \frac{\log{\left(x \right)}}{x}
f = x + log(x)/x
The graph of the function
02468-8-6-4-2-1010-100100
The domain of the function
The points at which the function is not precisely defined:
x1=0x_{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+log(x)x=0x + \frac{\log{\left(x \right)}}{x} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=eW(2)2x_{1} = e^{- \frac{W\left(2\right)}{2}}
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to x + log(x)/x.
log(0)0\frac{\log{\left(0 \right)}}{0}
The result:
f(0)=~f{\left(0 \right)} = \tilde{\infty}
sof doesn't intersect Y
Extrema of the function
In order to find the extrema, we need to solve the equation
ddxf(x)=0\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:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
the first derivative
1log(x)x2+1x2=01 - \frac{\log{\left(x \right)}}{x^{2}} + \frac{1}{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
d2dx2f(x)=0\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:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
2log(x)3x3=0\frac{2 \log{\left(x \right)} - 3}{x^{3}} = 0
Solve this equation
The roots of this equation
x1=e32x_{1} = e^{\frac{3}{2}}
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
x1=0x_{1} = 0

limx0(2log(x)3x3)=\lim_{x \to 0^-}\left(\frac{2 \log{\left(x \right)} - 3}{x^{3}}\right) = \infty
limx0+(2log(x)3x3)=\lim_{x \to 0^+}\left(\frac{2 \log{\left(x \right)} - 3}{x^{3}}\right) = -\infty
- the limits are not equal, so
x1=0x_{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
[e32,)\left[e^{\frac{3}{2}}, \infty\right)
Convex at the intervals
(,e32]\left(-\infty, e^{\frac{3}{2}}\right]
Vertical asymptotes
Have:
x1=0x_{1} = 0
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
limx(x+log(x)x)=\lim_{x \to -\infty}\left(x + \frac{\log{\left(x \right)}}{x}\right) = -\infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
limx(x+log(x)x)=\lim_{x \to \infty}\left(x + \frac{\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, divided by x at x->+oo and x ->-oo
limx(x+log(x)xx)=1\lim_{x \to -\infty}\left(\frac{x + \frac{\log{\left(x \right)}}{x}}{x}\right) = 1
Let's take the limit
so,
inclined asymptote equation on the left:
y=xy = x
limx(x+log(x)xx)=1\lim_{x \to \infty}\left(\frac{x + \frac{\log{\left(x \right)}}{x}}{x}\right) = 1
Let's take the limit
so,
inclined asymptote equation on the right:
y=xy = 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+log(x)x=xlog(x)xx + \frac{\log{\left(x \right)}}{x} = - x - \frac{\log{\left(- x \right)}}{x}
- No
x+log(x)x=x+log(x)xx + \frac{\log{\left(x \right)}}{x} = x + \frac{\log{\left(- x \right)}}{x}
- No
so, the function
not is
neither even, nor odd