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Plotting a function graph/ sin(x)+(1/2)sin(2x)

Graphing y = sin(x)+(1/2)sin(2x)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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                sin(2*x)
f(x) = sin(x) + --------
                   2
f(x)=sin⁡(x)+sin⁡(2x)2f{\left(x \right)} = \sin{\left(x \right)} + \frac{\sin{\left(2 x \right)}}{2} 
f = sin(x) + sin(2*x)/2
The graph of the function
02468-8-6-4-2-10102.5-2.5
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:
sin⁡(x)+sin⁡(2x)2=0\sin{\left(x \right)} + \frac{\sin{\left(2 x \right)}}{2} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=0x_{1} = 0
x2=πx_{2} = \pi
Numerical solution
x1=12.5663706143592x_{1} = 12.5663706143592
x2=97.389506033414x_{2} = 97.389506033414
x3=−28.2742611423571x_{3} = -28.2742611423571
x4=37.6991118430775x_{4} = 37.6991118430775
x5=72.2566292957295x_{5} = 72.2566292957295
x6=−91.1063173161218x_{6} = -91.1063173161218
x7=72.2564974285085x_{7} = 72.2564974285085
x8=−72.2565620594227x_{8} = -72.2565620594227
x9=−31.4159265358979x_{9} = -31.4159265358979
x10=0x_{10} = 0
x11=84.8228826659845x_{11} = 84.8228826659845
x12=−50.2654824574367x_{12} = -50.2654824574367
x13=−53.4071523808127x_{13} = -53.4071523808127
x14=6.28318530717959x_{14} = 6.28318530717959
x15=−94.2477796076938x_{15} = -94.2477796076938
x16=−69.1150383789755x_{16} = -69.1150383789755
x17=−59.6902836920888x_{17} = -59.6902836920888
x18=53.4072061236969x_{18} = 53.4072061236969
x19=−34.5573938477265x_{19} = -34.5573938477265
x20=65.9735385828884x_{20} = 65.9735385828884
x21=62.8318530717959x_{21} = 62.8318530717959
x22=50.2654824574367x_{22} = 50.2654824574367
x23=81.6814089933346x_{23} = 81.6814089933346
x24=100.530964914873x_{24} = 100.530964914873
x25=−87.9645943005142x_{25} = -87.9645943005142
x26=−65.9734547074718x_{26} = -65.9734547074718
x27=72.2566368440238x_{27} = 72.2566368440238
x28=−9.42485173622935x_{28} = -9.42485173622935
x29=−59.6902757594272x_{29} = -59.6902757594272
x30=78.5397496778866x_{30} = 78.5397496778866
x31=−15.7079741551755x_{31} = -15.7079741551755
x32=−56.5486677646163x_{32} = -56.5486677646163
x33=28.2742362262029x_{33} = 28.2742362262029
x34=−37.6991118430775x_{34} = -37.6991118430775
x35=−25.1327412287183x_{35} = -25.1327412287183
x36=−97.3894529845737x_{36} = -97.3894529845737
x37=−100.530964914873x_{37} = -100.530964914873
x38=−1083.8495084391x_{38} = -1083.8495084391
x39=−75.398223686155x_{39} = -75.398223686155
x40=18.8495559215388x_{40} = 18.8495559215388
x41=28.2743275355147x_{41} = 28.2743275355147
x42=34.557448949744x_{42} = 34.557448949744
x43=59.6903404916682x_{43} = 59.6903404916682
x44=65.9734548161256x_{44} = 65.9734548161256
x45=−6.28318530717959x_{45} = -6.28318530717959
x46=−3.14171741723949x_{46} = -3.14171741723949
x47=9.42490616313103x_{47} = 9.42490616313103
x48=−15.7080226365019x_{48} = -15.7080226365019
x49=56.5486677646163x_{49} = 56.5486677646163
x50=−43.9822971502571x_{50} = -43.9822971502571
x51=−47.1240173901594x_{51} = -47.1240173901594
x52=21.9912781084223x_{52} = 21.9912781084223
x53=15.7080397066029x_{53} = 15.7080397066029
x54=31.4159265358979x_{54} = 31.4159265358979
x55=94.2477796076938x_{55} = 94.2477796076938
x56=−12.5663706143592x_{56} = -12.5663706143592
x57=−78.5396939083992x_{57} = -78.5396939083992
x58=75.398223686155x_{58} = 75.398223686155
x59=40.8405826017537x_{59} = 40.8405826017537
x60=−21.9911516405744x_{60} = -21.9911516405744
x61=21.9911516419074x_{61} = 21.9911516419074
x62=−81.6814089933346x_{62} = -81.6814089933346
x63=43.9822971502571x_{63} = 43.9822971502571
x64=87.9645943005142x_{64} = 87.9645943005142
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to sin(x) + sin(2*x)/2.
sin⁡(0)+sin⁡(0⋅2)2\sin{\left(0 \right)} + \frac{\sin{\left(0 \cdot 2 \right)}}{2}
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
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
cos⁡(x)+cos⁡(2x)=0\cos{\left(x \right)} + \cos{\left(2 x \right)} = 0
Solve this equation
The roots of this equation
x1=−5π3x_{1} = - \frac{5 \pi}{3}
x2=−πx_{2} = - \pi
x3=−π3x_{3} = - \frac{\pi}{3}
x4=π3x_{4} = \frac{\pi}{3}
x5=πx_{5} = \pi
x6=5π3x_{6} = \frac{5 \pi}{3}
The values of the extrema at the points:
            ___ 
 -5*pi  3*\/ 3  
(-----, -------)
   3       4    

(-pi, 0)

            ___ 
 -pi   -3*\/ 3  
(----, --------)
  3       4     

         ___ 
 pi  3*\/ 3  
(--, -------)
 3      4    

(pi, 0)

            ___ 
 5*pi  -3*\/ 3  
(----, --------)
  3       4


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=−π3x_{1} = - \frac{\pi}{3}
x2=5π3x_{2} = \frac{5 \pi}{3}
Maxima of the function at points:
x2=−5π3x_{2} = - \frac{5 \pi}{3}
x2=π3x_{2} = \frac{\pi}{3}
Decreasing at intervals
[5π3,∞)\left[\frac{5 \pi}{3}, \infty\right)
Increasing at intervals
(−∞,−π3]∪[π3,5π3]\left(-\infty, - \frac{\pi}{3}\right] \cup \left[\frac{\pi}{3}, \frac{5 \pi}{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
−(sin⁡(x)+2sin⁡(2x))=0- (\sin{\left(x \right)} + 2 \sin{\left(2 x \right)}) = 0
Solve this equation
The roots of this equation
x1=0x_{1} = 0
x2=πx_{2} = \pi
x3=−ilog⁡(−14−15i4)x_{3} = - i \log{\left(- \frac{1}{4} - \frac{\sqrt{15} i}{4} \right)}
x4=−ilog⁡(−14+15i4)x_{4} = - i \log{\left(- \frac{1}{4} + \frac{\sqrt{15} i}{4} \right)}

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
[−π+atan⁡(15),0]∪[π−atan⁡(15),∞)\left[- \pi + \operatorname{atan}{\left(\sqrt{15} \right)}, 0\right] \cup \left[\pi - \operatorname{atan}{\left(\sqrt{15} \right)}, \infty\right)
Convex at the intervals
(−∞,−π+atan⁡(15)]\left(-\infty, - \pi + \operatorname{atan}{\left(\sqrt{15} \right)}\right]
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(sin⁡(x)+sin⁡(2x)2)=⟨−32,32⟩\lim_{x \to -\infty}\left(\sin{\left(x \right)} + \frac{\sin{\left(2 x \right)}}{2}\right) = \left\langle - \frac{3}{2}, \frac{3}{2}\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=⟨−32,32⟩y = \left\langle - \frac{3}{2}, \frac{3}{2}\right\rangle
lim⁡x→∞(sin⁡(x)+sin⁡(2x)2)=⟨−32,32⟩\lim_{x \to \infty}\left(\sin{\left(x \right)} + \frac{\sin{\left(2 x \right)}}{2}\right) = \left\langle - \frac{3}{2}, \frac{3}{2}\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=⟨−32,32⟩y = \left\langle - \frac{3}{2}, \frac{3}{2}\right\rangle
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of sin(x) + sin(2*x)/2, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(sin⁡(x)+sin⁡(2x)2x)=0\lim_{x \to -\infty}\left(\frac{\sin{\left(x \right)} + \frac{\sin{\left(2 x \right)}}{2}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(sin⁡(x)+sin⁡(2x)2x)=0\lim_{x \to \infty}\left(\frac{\sin{\left(x \right)} + \frac{\sin{\left(2 x \right)}}{2}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
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:
sin⁡(x)+sin⁡(2x)2=−sin⁡(x)−sin⁡(2x)2\sin{\left(x \right)} + \frac{\sin{\left(2 x \right)}}{2} = - \sin{\left(x \right)} - \frac{\sin{\left(2 x \right)}}{2}
- No
sin⁡(x)+sin⁡(2x)2=sin⁡(x)+sin⁡(2x)2\sin{\left(x \right)} + \frac{\sin{\left(2 x \right)}}{2} = \sin{\left(x \right)} + \frac{\sin{\left(2 x \right)}}{2}
- No
so, the function
not is
neither even, nor odd
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