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Graphing y = (x+exp(x))/(1+x)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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            x
       x + e 
f(x) = ------
       1 + x 
f(x)=x+exx+1f{\left(x \right)} = \frac{x + e^{x}}{x + 1}
f = (x + exp(x))/(x + 1)
The graph of the function
02468-8-6-4-2-10104000-2000
The domain of the function
The points at which the function is not precisely defined:
x1=1x_{1} = -1
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+exx+1=0\frac{x + e^{x}}{x + 1} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=W(1)x_{1} = - W\left(1\right)
Numerical solution
x1=0.567143290409784x_{1} = -0.567143290409784
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (x + exp(x))/(1 + x).
e01\frac{e^{0}}{1}
The result:
f(0)=1f{\left(0 \right)} = 1
The point:
(0, 1)
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
ex+1x+1x+ex(x+1)2=0\frac{e^{x} + 1}{x + 1} - \frac{x + e^{x}}{\left(x + 1\right)^{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
ex2(ex+1)x+1+2(x+ex)(x+1)2x+1=0\frac{e^{x} - \frac{2 \left(e^{x} + 1\right)}{x + 1} + \frac{2 \left(x + e^{x}\right)}{\left(x + 1\right)^{2}}}{x + 1} = 0
Solve this equation
The roots of this equation
x1=33415.8037497305x_{1} = -33415.8037497305
x2=28330.187841268x_{2} = -28330.187841268
x3=38501.422170149x_{3} = -38501.422170149
x4=23244.5760938158x_{4} = -23244.5760938158
x5=15616.1751697251x_{5} = -15616.1751697251
x6=30872.9954039089x_{6} = -30872.9954039089
x7=24092.177650919x_{7} = -24092.177650919
x8=0.483259389716171x_{8} = 0.483259389716171
x9=37653.818968837x_{9} = -37653.818968837
x10=19854.1721532085x_{10} = -19854.1721532085
x11=30025.392786181x_{11} = -30025.392786181
x12=26634.9833453947x_{12} = -26634.9833453947
x13=19006.5718930801x_{13} = -19006.5718930801
x14=29177.7902624869x_{14} = -29177.7902624869
x15=35958.6127125654x_{15} = -35958.6127125654
x16=18158.9720042001x_{16} = -18158.9720042001
x17=34263.4066755697x_{17} = -34263.4066755697
x18=20701.7727389755x_{18} = -20701.7727389755
x19=22396.9747395139x_{19} = -22396.9747395139
x20=13920.98050271x_{20} = -13920.98050271
x21=32568.2008920951x_{21} = -32568.2008920951
x22=39349.0254160413x_{22} = -39349.0254160413
x23=35111.0096646728x_{23} = -35111.0096646728
x24=41044.2320305184x_{24} = -41044.2320305184
x25=31720.5981081316x_{25} = -31720.5981081316
x26=25787.3812935264x_{26} = -25787.3812935264
x27=21549.3736119485x_{27} = -21549.3736119485
x28=17311.3725411133x_{28} = -17311.3725411133
x29=40196.6287036936x_{29} = -40196.6287036936
x30=41891.8353941377x_{30} = -41891.8353941377
x31=24939.779390143x_{31} = -24939.779390143
x32=13073.3845123319x_{32} = -13073.3845123319
x33=36806.2158151855x_{33} = -36806.2158151855
x34=16463.7735695997x_{34} = -16463.7735695997
x35=14768.5774399437x_{35} = -14768.5774399437
x36=27482.5855320072x_{36} = -27482.5855320072
x37=12225.7896664061x_{37} = -12225.7896664061
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=1x_{1} = -1

limx1(ex2(ex+1)x+1+2(x+ex)(x+1)2x+1)=\lim_{x \to -1^-}\left(\frac{e^{x} - \frac{2 \left(e^{x} + 1\right)}{x + 1} + \frac{2 \left(x + e^{x}\right)}{\left(x + 1\right)^{2}}}{x + 1}\right) = \infty
limx1+(ex2(ex+1)x+1+2(x+ex)(x+1)2x+1)=\lim_{x \to -1^+}\left(\frac{e^{x} - \frac{2 \left(e^{x} + 1\right)}{x + 1} + \frac{2 \left(x + e^{x}\right)}{\left(x + 1\right)^{2}}}{x + 1}\right) = -\infty
- the limits are not equal, so
x1=1x_{1} = -1
- 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
[0.483259389716171,)\left[0.483259389716171, \infty\right)
Convex at the intervals
(,0.483259389716171]\left(-\infty, 0.483259389716171\right]
Vertical asymptotes
Have:
x1=1x_{1} = -1
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
limx(x+exx+1)=1\lim_{x \to -\infty}\left(\frac{x + e^{x}}{x + 1}\right) = 1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=1y = 1
limx(x+exx+1)=\lim_{x \to \infty}\left(\frac{x + e^{x}}{x + 1}\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 + exp(x))/(1 + x), divided by x at x->+oo and x ->-oo
limx(x+exx(x+1))=0\lim_{x \to -\infty}\left(\frac{x + e^{x}}{x \left(x + 1\right)}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
limx(x+exx(x+1))=\lim_{x \to \infty}\left(\frac{x + e^{x}}{x \left(x + 1\right)}\right) = \infty
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
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+exx+1=x+ex1x\frac{x + e^{x}}{x + 1} = \frac{- x + e^{- x}}{1 - x}
- No
x+exx+1=x+ex1x\frac{x + e^{x}}{x + 1} = - \frac{- x + e^{- x}}{1 - x}
- No
so, the function
not is
neither even, nor odd