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

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

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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                        x
f(x) = (1 + x*(1 + x))*e
f(x)=(x(x+1)+1)exf{\left(x \right)} = \left(x \left(x + 1\right) + 1\right) e^{x} 
f = (x*(x + 1) + 1)*exp(x)
The graph of the function
02468-8-6-4-2-101002500000
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(x+1)+1)ex=0\left(x \left(x + 1\right) + 1\right) e^{x} = 0
Solve this equation
The points of intersection with the axis X:

Numerical solution
x1=−105.333005739691x_{1} = -105.333005739691
x2=−57.8505831390582x_{2} = -57.8505831390582
x3=−111.304489897731x_{3} = -111.304489897731
x4=−101.354206694808x_{4} = -101.354206694808
x5=−121.264081496849x_{5} = -121.264081496849
x6=−71.6086441089967x_{6} = -71.6086441089967
x7=−4135.94355451445x_{7} = -4135.94355451445
x8=−89.4314266145357x_{8} = -89.4314266145357
x9=−79.5176749819744x_{9} = -79.5176749819744
x10=−81.4983526999876x_{10} = -81.4983526999876
x11=−40.5801277728577x_{11} = -40.5801277728577
x12=−48.1529659582005x_{12} = -48.1529659582005
x13=−113.295754144651x_{13} = -113.295754144651
x14=−95.3899462845119x_{14} = -95.3899462845119
x15=−36.9291459649712x_{15} = -36.9291459649712
x16=−531.837727606964x_{16} = -531.837727606964
x17=−109.313594149021x_{17} = -109.313594149021
x18=−61.7664973306176x_{18} = -61.7664973306176
x19=−63.72969387164x_{19} = -63.72969387164
x20=−77.5382332700649x_{20} = -77.5382332700649
x21=−52.0117442489976x_{21} = -52.0117442489976
x22=−69.6355644697641x_{22} = -69.6355644697641
x23=−46.2378882480841x_{23} = -46.2378882480841
x24=−97.3774570975963x_{24} = -97.3774570975963
x25=−50.0781629625302x_{25} = -50.0781629625302
x26=−42.4478982713749x_{26} = -42.4478982713749
x27=−35.1676235533282x_{27} = -35.1676235533282
x28=−65.6958216263569x_{28} = -65.6958216263569
x29=−53.9523555154393x_{29} = -53.9523555154393
x30=−91.4168785530278x_{30} = -91.4168785530278
x31=−44.3351933877829x_{31} = -44.3351933877829
x32=−117.279301972525x_{32} = -117.279301972525
x33=−119.271546597134x_{33} = -119.271546597134
x34=−85.4629939239387x_{34} = -85.4629939239387
x35=−67.6645413261944x_{35} = -67.6645413261944
x36=−75.5601507367764x_{36} = -75.5601507367764
x37=−107.323090768864x_{37} = -107.323090768864
x38=−93.4030701679791x_{38} = -93.4030701679791
x39=−83.4801577602678x_{39} = -83.4801577602678
x40=−73.5835675835769x_{40} = -73.5835675835769
x41=−51.1656821821462x_{41} = -51.1656821821462
x42=−103.343367395508x_{42} = -103.343367395508
x43=−99.3655575325085x_{43} = -99.3655575325085
x44=−55.8989228973151x_{44} = -55.8989228973151
x45=−38.7376998275312x_{45} = -38.7376998275312
x46=−87.4467756438286x_{46} = -87.4467756438286
x47=−59.8066338460799x_{47} = -59.8066338460799
x48=−115.287364920786x_{48} = -115.287364920786
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (1 + x*(1 + x))*exp(x).
e0e^{0}
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
(2x+1)ex+(x(x+1)+1)ex=0\left(2 x + 1\right) e^{x} + \left(x \left(x + 1\right) + 1\right) e^{x} = 0
Solve this equation
The roots of this equation
x1=−2x_{1} = -2
x2=−1x_{2} = -1
The values of the extrema at the points:
        -2 
(-2, 3*e  )

      -1 
(-1, 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:
Minima of the function at points:
x1=−1x_{1} = -1
Maxima of the function at points:
x1=−2x_{1} = -2
Decreasing at intervals
(−∞,−2]∪[−1,∞)\left(-\infty, -2\right] \cup \left[-1, \infty\right)
Increasing at intervals
[−2,−1]\left[-2, -1\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(x+1)+4x+5)ex=0\left(x \left(x + 1\right) + 4 x + 5\right) e^{x} = 0
Solve this equation
The roots of this equation
x1=−52−52x_{1} = - \frac{5}{2} - \frac{\sqrt{5}}{2}
x2=−52+52x_{2} = - \frac{5}{2} + \frac{\sqrt{5}}{2}

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
(−∞,−52−52]∪[−52+52,∞)\left(-\infty, - \frac{5}{2} - \frac{\sqrt{5}}{2}\right] \cup \left[- \frac{5}{2} + \frac{\sqrt{5}}{2}, \infty\right)
Convex at the intervals
[−52−52,−52+52]\left[- \frac{5}{2} - \frac{\sqrt{5}}{2}, - \frac{5}{2} + \frac{\sqrt{5}}{2}\right]
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞((x(x+1)+1)ex)=0\lim_{x \to -\infty}\left(\left(x \left(x + 1\right) + 1\right) e^{x}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞((x(x+1)+1)ex)=∞\lim_{x \to \infty}\left(\left(x \left(x + 1\right) + 1\right) e^{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 (1 + x*(1 + x))*exp(x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞((x(x+1)+1)exx)=0\lim_{x \to -\infty}\left(\frac{\left(x \left(x + 1\right) + 1\right) e^{x}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞((x(x+1)+1)exx)=∞\lim_{x \to \infty}\left(\frac{\left(x \left(x + 1\right) + 1\right) e^{x}}{x}\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(x+1)+1)ex=(−x(1−x)+1)e−x\left(x \left(x + 1\right) + 1\right) e^{x} = \left(- x \left(1 - x\right) + 1\right) e^{- x}
- No
(x(x+1)+1)ex=−(−x(1−x)+1)e−x\left(x \left(x + 1\right) + 1\right) e^{x} = - \left(- x \left(1 - x\right) + 1\right) e^{- x}
- No
so, the function
not is
neither even, nor odd
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