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

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

v

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
Solution is not found,
it's possible that the graph doesn't intersect the axis X
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).
0e0+11\frac{0 e^{0} + 1}{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
The roots of this equation
x1=0x_{1} = 0
The values of the extrema at the points:
(0, 1)


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:
Minima of the function at points:
x1=0x_{1} = 0
The function has no maxima
Decreasing at intervals
[0,∞)\left[0, \infty\right)
Increasing at intervals
(−∞,0]\left(-\infty, 0\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
(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=−33415.8037497305x_{1} = -33415.8037497305
x2=−28330.187841268x_{2} = -28330.187841268
x3=−38501.422170149x_{3} = -38501.422170149
x4=−23244.5760938158x_{4} = -23244.5760938158
x5=−30872.9954039089x_{5} = -30872.9954039089
x6=−19854.1721532085x_{6} = -19854.1721532085
x7=−24092.177650919x_{7} = -24092.177650919
x8=−18158.9720042003x_{8} = -18158.9720042003
x9=−37653.818968837x_{9} = -37653.818968837
x10=−30025.392786181x_{10} = -30025.392786181
x11=−26634.9833453947x_{11} = -26634.9833453947
x12=−17311.3725411143x_{12} = -17311.3725411143
x13=−29177.7902624869x_{13} = -29177.7902624869
x14=−16463.7735696055x_{14} = -16463.7735696055
x15=−35958.6127125654x_{15} = -35958.6127125654
x16=−34263.4066755697x_{16} = -34263.4066755697
x17=−22396.9747395139x_{17} = -22396.9747395139
x18=−19006.5718930801x_{18} = -19006.5718930801
x19=−32568.2008920951x_{19} = -32568.2008920951
x20=−39349.0254160413x_{20} = -39349.0254160413
x21=−35111.0096646728x_{21} = -35111.0096646728
x22=−20701.7727389755x_{22} = -20701.7727389755
x23=−13073.3845178082x_{23} = -13073.3845178082
x24=−12225.7896953671x_{24} = -12225.7896953671
x25=−41044.2320305184x_{25} = -41044.2320305184
x26=−31720.5981081316x_{26} = -31720.5981081316
x27=−25787.3812935264x_{27} = -25787.3812935264
x28=−21549.3736119485x_{28} = -21549.3736119485
x29=−15616.1751697581x_{29} = -15616.1751697581
x30=−14768.577440128x_{30} = -14768.577440128
x31=−40196.6287036936x_{31} = -40196.6287036936
x32=−13920.9805037239x_{32} = -13920.9805037239
x33=−41891.8353941377x_{33} = -41891.8353941377
x34=−24939.779390143x_{34} = -24939.779390143
x35=−36806.2158151855x_{35} = -36806.2158151855
x36=−27482.5855320072x_{36} = -27482.5855320072
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:
Have no bends at the whole real axis
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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