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  • Graphing y =:
  • (x+1)(x-2)^2
  • (x-1)^2*(x+2)
  • (9x^2-1)/x
  • 9^(1/(x-3))

  • Identical expressions

  • e^(two *x)*(x+ one)
  • e to the power of (2 multiply by x) multiply by (x plus 1)
  • e to the power of (two multiply by x) multiply by (x plus one)
  • e(2*x)*(x+1)
  • e2*x*x+1
  • e^(2x)(x+1)
  • e(2x)(x+1)
  • e2xx+1
  • e^2xx+1

  • Similar expressions

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

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

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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