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

Graphing y = (x+1)*e^(2*x)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
                2*x
f(x) = (x + 1)*E
f(x)=e2x(x+1)f{\left(x \right)} = e^{2 x} \left(x + 1\right) 
f = E^(2*x)*(x + 1)
The graph of the function
02468-8-6-4-2-1010-500000000010000000000
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:
e2x(x+1)=0e^{2 x} \left(x + 1\right) = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−1x_{1} = -1
Numerical solution
x1=−50.487646689892x_{1} = -50.487646689892
x2=−46.5000993631074x_{2} = -46.5000993631074
x3=−66.454745139734x_{3} = -66.454745139734
x4=−40.5244692579679x_{4} = -40.5244692579679
x5=−104.420095524533x_{5} = -104.420095524533
x6=−38.5347052738485x_{6} = -38.5347052738485
x7=−62.4611748339881x_{7} = -62.4611748339881
x8=−102.421228908762x_{8} = -102.421228908762
x9=−76.4419310233885x_{9} = -76.4419310233885
x10=−70.4491316757143x_{10} = -70.4491316757143
x11=−15.2083025251737x_{11} = -15.2083025251737
x12=−68.4518464300967x_{12} = -68.4518464300967
x13=−28.6156935676912x_{13} = -28.6156935676912
x14=−42.5154118585931x_{14} = -42.5154118585931
x15=−88.4307270314252x_{15} = -88.4307270314252
x16=−54.4773187883893x_{16} = -54.4773187883893
x17=−90.429174285909x_{17} = -90.429174285909
x18=−16.9806772001439x_{18} = -16.9806772001439
x19=−96.4249341851903x_{19} = -96.4249341851903
x20=−64.4578471989128x_{20} = -64.4578471989128
x21=−36.5463680381357x_{21} = -36.5463680381357
x22=−82.4358814217583x_{22} = -82.4358814217583
x23=−1x_{23} = -1
x24=−20.7755860151928x_{24} = -20.7755860151928
x25=−92.4276945075164x_{25} = -92.4276945075164
x26=−60.464753641239x_{26} = -60.464753641239
x27=−106.419007861607x_{27} = -106.419007861607
x28=−34.5597813171015x_{28} = -34.5597813171015
x29=−78.439801011308x_{29} = -78.439801011308
x30=−24.6757451532018x_{30} = -24.6757451532018
x31=−86.432358283505x_{31} = -86.432358283505
x32=−94.4262826664454x_{32} = -94.4262826664454
x33=−80.437787586974x_{33} = -80.437787586974
x34=−108.417963206245x_{34} = -108.417963206245
x35=−30.5937393694414x_{35} = -30.5937393694414
x36=−58.4686132524805x_{36} = -58.4686132524805
x37=−22.7185017258284x_{37} = -22.7185017258284
x38=−48.4935682921138x_{38} = -48.4935682921138
x39=−84.4340741575283x_{39} = -84.4340741575283
x40=−98.4236448887784x_{40} = -98.4236448887784
x41=−74.4441880676345x_{41} = -74.4441880676345
x42=−26.6424295706756x_{42} = -26.6424295706756
x43=−110.416959055435x_{43} = -110.416959055435
x44=−100.42241096144x_{44} = -100.42241096144
x45=−44.5073396063464x_{45} = -44.5073396063464
x46=−56.4727881654018x_{46} = -56.4727881654018
x47=−32.5753760558308x_{47} = -32.5753760558308
x48=−52.482252767615x_{48} = -52.482252767615
x49=−72.4465838765021x_{49} = -72.4465838765021
x50=−18.8562161179393x_{50} = -18.8562161179393
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (x + 1)*E^(2*x).
e0⋅2e^{0 \cdot 2}
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
2(x+1)e2x+e2x=02 \left(x + 1\right) e^{2 x} + e^{2 x} = 0
Solve this equation
The roots of this equation
x1=−32x_{1} = - \frac{3}{2}
The values of the extrema at the points:
         -3  
       -e    
(-3/2, -----)
         2


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=−32x_{1} = - \frac{3}{2}
The function has no maxima
Decreasing at intervals
[−32,∞)\left[- \frac{3}{2}, \infty\right)
Increasing at intervals
(−∞,−32]\left(-\infty, - \frac{3}{2}\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
4(x+2)e2x=04 \left(x + 2\right) e^{2 x} = 0
Solve this equation
The roots of this equation
x1=−2x_{1} = -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
[−2,∞)\left[-2, \infty\right)
Convex at the intervals
(−∞,−2]\left(-\infty, -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→−∞(e2x(x+1))=0\lim_{x \to -\infty}\left(e^{2 x} \left(x + 1\right)\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(e2x(x+1))=∞\lim_{x \to \infty}\left(e^{2 x} \left(x + 1\right)\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 + 1)*E^(2*x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞((x+1)e2xx)=0\lim_{x \to -\infty}\left(\frac{\left(x + 1\right) e^{2 x}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞((x+1)e2xx)=∞\lim_{x \to \infty}\left(\frac{\left(x + 1\right) e^{2 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:
e2x(x+1)=(1−x)e−2xe^{2 x} \left(x + 1\right) = \left(1 - x\right) e^{- 2 x}
- No
e2x(x+1)=−(1−x)e−2xe^{2 x} \left(x + 1\right) = - \left(1 - x\right) e^{- 2 x}
- No
so, the function
not is
neither even, nor odd
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