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Plotting a function graph/ ln*x/(x+2)+1

Graphing y = ln*x/(x+2)+1

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
       log(x)    
f(x) = ------ + 1
       x + 2
f(x)=1+log⁡(x)x+2f{\left(x \right)} = 1 + \frac{\log{\left(x \right)}}{x + 2} 
f = 1 + log(x)/(x + 2)
The graph of the function
02468-8-6-4-2-10102-2
The domain of the function
The points at which the function is not precisely defined:
x1=−2x_{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:
1+log⁡(x)x+2=01 + \frac{\log{\left(x \right)}}{x + 2} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=W(e−2)x_{1} = W\left(e^{-2}\right)
Numerical solution
x1=0.120028238987641x_{1} = 0.120028238987641
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to log(x)/(x + 2) + 1.
log⁡(0)2+1\frac{\log{\left(0 \right)}}{2} + 1
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
−log⁡(x)(x+2)2+1x(x+2)=0- \frac{\log{\left(x \right)}}{\left(x + 2\right)^{2}} + \frac{1}{x \left(x + 2\right)} = 0
Solve this equation
The roots of this equation
x1=eW(2e)+1x_{1} = e^{W\left(\frac{2}{e}\right) + 1}
The values of the extrema at the points:
       /   -1\              /   -1\   
  1 + W\2*e  /         1 + W\2*e  /   
(e           , 1 + -----------------)
                              /   -1\ 
                         1 + W\2*e  / 
                    2 + e


Intervals of increase and decrease of the function:
Let's find intervals where the function increases and decreases, as well as minima and maxima of the function, for this let's look how the function behaves itself in the extremas and at the slightest deviation from:
The function has no minima
Maxima of the function at points:
x1=eW(2e)+1x_{1} = e^{W\left(\frac{2}{e}\right) + 1}
Decreasing at intervals
(−∞,eW(2e)+1]\left(-\infty, e^{W\left(\frac{2}{e}\right) + 1}\right]
Increasing at intervals
[eW(2e)+1,∞)\left[e^{W\left(\frac{2}{e}\right) + 1}, \infty\right)
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)(x+2)2−2x(x+2)−1x2x+2=0\frac{\frac{2 \log{\left(x \right)}}{\left(x + 2\right)^{2}} - \frac{2}{x \left(x + 2\right)} - \frac{1}{x^{2}}}{x + 2} = 0
Solve this equation
The roots of this equation
x1=24294.3612637142x_{1} = 24294.3612637142
x2=55296.0175240638x_{2} = 55296.0175240638
x3=37615.0545701149x_{3} = 37615.0545701149
x4=29835.2864925341x_{4} = 29835.2864925341
x5=36504.5395380226x_{5} = 36504.5395380226
x6=42051.3926649508x_{6} = 42051.3926649508
x7=48686.0781758374x_{7} = 48686.0781758374
x8=26505.142019564x_{8} = 26505.142019564
x9=45371.8340832042x_{9} = 45371.8340832042
x10=54196.0466546917x_{10} = 54196.0466546917
x11=59689.4020575102x_{11} = 59689.4020575102
x12=43158.8847440741x_{12} = 43158.8847440741
x13=44265.7017725604x_{13} = 44265.7017725604
x14=51994.1010647193x_{14} = 51994.1010647193
x15=35393.5551813782x_{15} = 35393.5551813782
x16=57493.9931924793x_{16} = 57493.9931924793
x17=46477.2757599979x_{17} = 46477.2757599979
x18=34282.1802372611x_{18} = 34282.1802372611
x19=33170.5159341437x_{19} = 33170.5159341437
x20=40943.2398361897x_{20} = 40943.2398361897
x21=53095.4101884672x_{21} = 53095.4101884672
x22=27613.9915244101x_{22} = 27613.9915244101
x23=60786.1652576388x_{23} = 60786.1652576388
x24=30946.8770750174x_{24} = 30946.8770750174
x25=25398.3042276186x_{25} = 25398.3042276186
x26=28724.2019017552x_{26} = 28724.2019017552
x27=39834.446294619x_{27} = 39834.446294619
x28=58592.0142527443x_{28} = 58592.0142527443
x29=56395.3303766332x_{29} = 56395.3303766332
x30=32058.6923765109x_{30} = 32058.6923765109
x31=47582.0239274937x_{31} = 47582.0239274937
x32=38725.0391970573x_{32} = 38725.0391970573
x33=7.75850139245233x_{33} = 7.75850139245233
x34=49789.4400935317x_{34} = 49789.4400935317
x35=50892.1128862442x_{35} = 50892.1128862442
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=−2x_{1} = -2

lim⁡x→−2−(2log⁡(x)(x+2)2−2x(x+2)−1x2x+2)=−∞sign⁡(1.38629436111989+2iπ)\lim_{x \to -2^-}\left(\frac{\frac{2 \log{\left(x \right)}}{\left(x + 2\right)^{2}} - \frac{2}{x \left(x + 2\right)} - \frac{1}{x^{2}}}{x + 2}\right) = - \infty \operatorname{sign}{\left(1.38629436111989 + 2 i \pi \right)}
lim⁡x→−2+(2log⁡(x)(x+2)2−2x(x+2)−1x2x+2)=∞sign⁡(1.38629436111989+2iπ)\lim_{x \to -2^+}\left(\frac{\frac{2 \log{\left(x \right)}}{\left(x + 2\right)^{2}} - \frac{2}{x \left(x + 2\right)} - \frac{1}{x^{2}}}{x + 2}\right) = \infty \operatorname{sign}{\left(1.38629436111989 + 2 i \pi \right)}
- the limits are not equal, so
x1=−2x_{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
[7.75850139245233,∞)\left[7.75850139245233, \infty\right)
Convex at the intervals
(−∞,7.75850139245233]\left(-\infty, 7.75850139245233\right]
Vertical asymptotes
Have:
x1=−2x_{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→−∞(1+log⁡(x)x+2)=1\lim_{x \to -\infty}\left(1 + \frac{\log{\left(x \right)}}{x + 2}\right) = 1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=1y = 1
lim⁡x→∞(1+log⁡(x)x+2)=1\lim_{x \to \infty}\left(1 + \frac{\log{\left(x \right)}}{x + 2}\right) = 1
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=1y = 1
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of log(x)/(x + 2) + 1, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(1+log⁡(x)x+2x)=0\lim_{x \to -\infty}\left(\frac{1 + \frac{\log{\left(x \right)}}{x + 2}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(1+log⁡(x)x+2x)=0\lim_{x \to \infty}\left(\frac{1 + \frac{\log{\left(x \right)}}{x + 2}}{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:
1+log⁡(x)x+2=1+log⁡(−x)2−x1 + \frac{\log{\left(x \right)}}{x + 2} = 1 + \frac{\log{\left(- x \right)}}{2 - x}
- No
1+log⁡(x)x+2=−1−log⁡(−x)2−x1 + \frac{\log{\left(x \right)}}{x + 2} = -1 - \frac{\log{\left(- x \right)}}{2 - x}
- No
so, the function
not is
neither even, nor odd
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