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Plotting a function graph/ (1/3)(x-1)e^(3x+1)

Graphing y = (1/3)(x-1)e^(3x+1)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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       x - 1  3*x + 1
f(x) = -----*E       
         3
f(x)=e3x+1x−13f{\left(x \right)} = e^{3 x + 1} \frac{x - 1}{3} 
f = E^(3*x + 1)*((x - 1)/3)
The graph of the function
0-80-60-40-2020406080-100100-2e1322e132
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:
e3x+1x−13=0e^{3 x + 1} \frac{x - 1}{3} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=1x_{1} = 1
Numerical solution
x1=−15.0684532551434x_{1} = -15.0684532551434
x2=−100.891464090763x_{2} = -100.891464090763
x3=−96.8924798052928x_{3} = -96.8924798052928
x4=−60.9079523700679x_{4} = -60.9079523700679
x5=−106.890087328775x_{5} = -106.890087328775
x6=−92.8935860327674x_{6} = -92.8935860327674
x7=−70.902004841254x_{7} = -70.902004841254
x8=−32.9459328554122x_{8} = -32.9459328554122
x9=−13.1117508530767x_{9} = -13.1117508530767
x10=−66.9041598631251x_{10} = -66.9041598631251
x11=−24.9748914520213x_{11} = -24.9748914520213
x12=−68.9030495676499x_{12} = -68.9030495676499
x13=−76.8992106900462x_{13} = -76.8992106900462
x14=−74.9000901450155x_{14} = -74.9000901450155
x15=−54.9126249353434x_{15} = -54.9126249353434
x16=−22.9858493249115x_{16} = -22.9858493249115
x17=−88.8947954413616x_{17} = -88.8947954413616
x18=−48.918524600458x_{18} = -48.918524600458
x19=−17.0384287747027x_{19} = -17.0384287747027
x20=−56.9109514050207x_{20} = -56.9109514050207
x21=−34.9409892536895x_{21} = -34.9409892536895
x22=−46.9208513568803x_{22} = -46.9208513568803
x23=−102.8909867729x_{23} = -102.8909867729
x24=−42.9262089003892x_{24} = -42.9262089003892
x25=−26.9658032003852x_{25} = -26.9658032003852
x26=−62.9066035286623x_{26} = -62.9066035286623
x27=−36.9366344904773x_{27} = -36.9366344904773
x28=−82.8968369889989x_{28} = -82.8968369889989
x29=−44.9234015037324x_{29} = -44.9234015037324
x30=−20.9993264203452x_{30} = -20.9993264203452
x31=−94.8930208649533x_{31} = -94.8930208649533
x32=−28.9581416011761x_{32} = -28.9581416011761
x33=−98.8919613429433x_{33} = -98.8919613429433
x34=1x_{34} = 1
x35=−50.9163930751628x_{35} = -50.9163930751628
x36=−52.9144331692945x_{36} = -52.9144331692945
x37=−30.9515939059262x_{37} = -30.9515939059262
x38=−19.0163171160267x_{38} = -19.0163171160267
x39=−90.8941769573876x_{39} = -90.8941769573876
x40=−104.89052821385x_{40} = -104.89052821385
x41=−38.9327690324926x_{41} = -38.9327690324926
x42=−72.9010200386809x_{42} = -72.9010200386809
x43=−86.895443459527x_{43} = -86.895443459527
x44=−11.1803054802931x_{44} = -11.1803054802931
x45=−84.8961231801092x_{45} = -84.8961231801092
x46=−58.9093980525525x_{46} = -58.9093980525525
x47=−80.8975875177835x_{47} = -80.8975875177835
x48=−40.929314628757x_{48} = -40.929314628757
x49=−78.8983776762272x_{49} = -78.8983776762272
x50=−64.9053421056098x_{50} = -64.9053421056098
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to ((x - 1)/3)*E^(3*x + 1).
−13e0⋅3+1\frac{-1}{3} e^{0 \cdot 3 + 1}
The result:
f(0)=−e3f{\left(0 \right)} = - \frac{e}{3}
The point:
(0, -E/3)
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
(x−1)e3x+1+e3x+13=0\left(x - 1\right) e^{3 x + 1} + \frac{e^{3 x + 1}}{3} = 0
Solve this equation
The roots of this equation
x1=23x_{1} = \frac{2}{3}
The values of the extrema at the points:
        3  
      -e   
(2/3, ----)
       9


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=23x_{1} = \frac{2}{3}
The function has no maxima
Decreasing at intervals
[23,∞)\left[\frac{2}{3}, \infty\right)
Increasing at intervals
(−∞,23]\left(-\infty, \frac{2}{3}\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
(3x−1)e3x+1=0\left(3 x - 1\right) e^{3 x + 1} = 0
Solve this equation
The roots of this equation
x1=13x_{1} = \frac{1}{3}

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
[13,∞)\left[\frac{1}{3}, \infty\right)
Convex at the intervals
(−∞,13]\left(-\infty, \frac{1}{3}\right]
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(e3x+1x−13)=0\lim_{x \to -\infty}\left(e^{3 x + 1} \frac{x - 1}{3}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(e3x+1x−13)=∞\lim_{x \to \infty}\left(e^{3 x + 1} \frac{x - 1}{3}\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)/3)*E^(3*x + 1), divided by x at x->+oo and x ->-oo
lim⁡x→−∞((x−1)e3x+13x)=0\lim_{x \to -\infty}\left(\frac{\left(x - 1\right) e^{3 x + 1}}{3 x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞((x−1)e3x+13x)=∞\lim_{x \to \infty}\left(\frac{\left(x - 1\right) e^{3 x + 1}}{3 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:
e3x+1x−13=(−x3−13)e1−3xe^{3 x + 1} \frac{x - 1}{3} = \left(- \frac{x}{3} - \frac{1}{3}\right) e^{1 - 3 x}
- No
e3x+1x−13=−(−x3−13)e1−3xe^{3 x + 1} \frac{x - 1}{3} = - \left(- \frac{x}{3} - \frac{1}{3}\right) e^{1 - 3 x}
- No
so, the function
not is
neither even, nor odd
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