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Plotting a function graph/ (-1+x*exp(x))/(1+x)

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

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
               x
       -1 + x*e 
f(x) = ---------
         1 + x
f(x)=xex−1x+1f{\left(x \right)} = \frac{x e^{x} - 1}{x + 1} 
f = (x*exp(x) - 1)/(x + 1)
The graph of the function
02468-8-6-4-2-101040000-20000
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:
xex−1x+1=0\frac{x e^{x} - 1}{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 (-1 + x*exp(x))/(1 + x).
−1+0e01\frac{-1 + 0 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
xex+exx+1−xex−1(x+1)2=0\frac{x e^{x} + e^{x}}{x + 1} - \frac{x e^{x} - 1}{\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
(x+2)ex−2ex+2(xex−1)(x+1)2x+1=0\frac{\left(x + 2\right) e^{x} - 2 e^{x} + \frac{2 \left(x e^{x} - 1\right)}{\left(x + 1\right)^{2}}}{x + 1} = 0
Solve this equation
The roots of this equation
x1=−25787.3812935264x_{1} = -25787.3812935264
x2=−30872.9954039089x_{2} = -30872.9954039089
x3=−35111.0096646728x_{3} = -35111.0096646728
x4=−36806.2158151855x_{4} = -36806.2158151855
x5=−28330.187841268x_{5} = -28330.187841268
x6=−24939.779390143x_{6} = -24939.779390143
x7=−26634.9833453947x_{7} = -26634.9833453947
x8=−35958.6127125654x_{8} = -35958.6127125654
x9=−19854.1721532085x_{9} = -19854.1721532085
x10=−14768.5774397478x_{10} = -14768.5774397478
x11=−17311.3725411122x_{11} = -17311.3725411122
x12=−13073.3845064641x_{12} = -13073.3845064641
x13=−24092.177650919x_{13} = -24092.177650919
x14=−32568.2008920951x_{14} = -32568.2008920951
x15=−22396.9747395139x_{15} = -22396.9747395139
x16=−34263.4066755697x_{16} = -34263.4066755697
x17=−20701.7727389755x_{17} = -20701.7727389755
x18=−30025.392786181x_{18} = -30025.392786181
x19=−27482.5855320072x_{19} = -27482.5855320072
x20=0.361524883587661x_{20} = 0.361524883587661
x21=−19006.57189308x_{21} = -19006.57189308
x22=−29177.7902624869x_{22} = -29177.7902624869
x23=−15616.1751696903x_{23} = -15616.1751696903
x24=−33415.8037497305x_{24} = -33415.8037497305
x25=−41891.8353941377x_{25} = -41891.8353941377
x26=−41044.2320305184x_{26} = -41044.2320305184
x27=−23244.5760938158x_{27} = -23244.5760938158
x28=−21549.3736119485x_{28} = -21549.3736119485
x29=−38501.422170149x_{29} = -38501.422170149
x30=−37653.818968837x_{30} = -37653.818968837
x31=−16463.7735695936x_{31} = -16463.7735695936
x32=−12225.7896352151x_{32} = -12225.7896352151
x33=−18158.9720041999x_{33} = -18158.9720041999
x34=−40196.6287036936x_{34} = -40196.6287036936
x35=−39349.0254160413x_{35} = -39349.0254160413
x36=−13920.9805016283x_{36} = -13920.9805016283
x37=−31720.5981081316x_{37} = -31720.5981081316
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

lim⁡x→−1−((x+2)ex−2ex+2(xex−1)(x+1)2x+1)=∞\lim_{x \to -1^-}\left(\frac{\left(x + 2\right) e^{x} - 2 e^{x} + \frac{2 \left(x e^{x} - 1\right)}{\left(x + 1\right)^{2}}}{x + 1}\right) = \infty
lim⁡x→−1+((x+2)ex−2ex+2(xex−1)(x+1)2x+1)=−∞\lim_{x \to -1^+}\left(\frac{\left(x + 2\right) e^{x} - 2 e^{x} + \frac{2 \left(x e^{x} - 1\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
[0.361524883587661,∞)\left[0.361524883587661, \infty\right)
Convex at the intervals
(−∞,0.361524883587661]\left(-\infty, 0.361524883587661\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
lim⁡x→−∞(xex−1x+1)=0\lim_{x \to -\infty}\left(\frac{x e^{x} - 1}{x + 1}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(xex−1x+1)=∞\lim_{x \to \infty}\left(\frac{x e^{x} - 1}{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 (-1 + x*exp(x))/(1 + x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(xex−1x(x+1))=0\lim_{x \to -\infty}\left(\frac{x e^{x} - 1}{x \left(x + 1\right)}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(xex−1x(x+1))=∞\lim_{x \to \infty}\left(\frac{x e^{x} - 1}{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:
xex−1x+1=−xe−x−11−x\frac{x e^{x} - 1}{x + 1} = \frac{- x e^{- x} - 1}{1 - x}
- No
xex−1x+1=−−xe−x−11−x\frac{x e^{x} - 1}{x + 1} = - \frac{- x e^{- x} - 1}{1 - x}
- No
so, the function
not is
neither even, nor odd
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