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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=28.2743275355147x_{1} = 28.2743275355147
x2=−15.7079741551755x_{2} = -15.7079741551755
x3=−75.398223686155x_{3} = -75.398223686155
x4=−31.4159265358979x_{4} = -31.4159265358979
x5=−91.1063173161218x_{5} = -91.1063173161218
x6=87.9645943005142x_{6} = 87.9645943005142
x7=−87.9645943005142x_{7} = -87.9645943005142
x8=15.7080397066029x_{8} = 15.7080397066029
x9=6.28318530717959x_{9} = 6.28318530717959
x10=−59.6902836920888x_{10} = -59.6902836920888
x11=100.530964914873x_{11} = 100.530964914873
x12=−59.6902757594272x_{12} = -59.6902757594272
x13=65.9734548161256x_{13} = 65.9734548161256
x14=62.8318530717959x_{14} = 62.8318530717959
x15=−53.4071523808127x_{15} = -53.4071523808127
x16=97.389506033414x_{16} = 97.389506033414
x17=−69.1150383789755x_{17} = -69.1150383789755
x18=28.2742362262029x_{18} = 28.2742362262029
x19=12.5663706143592x_{19} = 12.5663706143592
x20=94.2477796076938x_{20} = 94.2477796076938
x21=31.4159265358979x_{21} = 31.4159265358979
x22=−37.6991118430775x_{22} = -37.6991118430775
x23=−21.9911516405744x_{23} = -21.9911516405744
x24=−47.1240173901594x_{24} = -47.1240173901594
x25=−97.3894529845737x_{25} = -97.3894529845737
x26=−1083.8495084391x_{26} = -1083.8495084391
x27=−56.5486677646163x_{27} = -56.5486677646163
x28=81.6814089933346x_{28} = 81.6814089933346
x29=−94.2477796076938x_{29} = -94.2477796076938
x30=43.9822971502571x_{30} = 43.9822971502571
x31=−34.5573938477265x_{31} = -34.5573938477265
x32=21.9911516419074x_{32} = 21.9911516419074
x33=72.2566292957295x_{33} = 72.2566292957295
x34=59.6903404916682x_{34} = 59.6903404916682
x35=−3.14171741723949x_{35} = -3.14171741723949
x36=78.5397496778866x_{36} = 78.5397496778866
x37=72.2566368440238x_{37} = 72.2566368440238
x38=34.557448949744x_{38} = 34.557448949744
x39=56.5486677646163x_{39} = 56.5486677646163
x40=−100.530964914873x_{40} = -100.530964914873
x41=−28.2742611423571x_{41} = -28.2742611423571
x42=9.42490616313103x_{42} = 9.42490616313103
x43=84.8228826659845x_{43} = 84.8228826659845
x44=−15.7080226365019x_{44} = -15.7080226365019
x45=18.8495559215388x_{45} = 18.8495559215388
x46=−65.9734547074718x_{46} = -65.9734547074718
x47=0x_{47} = 0
x48=−43.9822971502571x_{48} = -43.9822971502571
x49=−72.2565620594227x_{49} = -72.2565620594227
x50=−12.5663706143592x_{50} = -12.5663706143592
x51=65.9735385828884x_{51} = 65.9735385828884
x52=40.8405826017537x_{52} = 40.8405826017537
x53=−6.28318530717959x_{53} = -6.28318530717959
x54=75.398223686155x_{54} = 75.398223686155
x55=−9.42485173622935x_{55} = -9.42485173622935
x56=−50.2654824574367x_{56} = -50.2654824574367
x57=−81.6814089933346x_{57} = -81.6814089933346
x58=50.2654824574367x_{58} = 50.2654824574367
x59=−78.5396939083992x_{59} = -78.5396939083992
x60=37.6991118430775x_{60} = 37.6991118430775
x61=−25.1327412287183x_{61} = -25.1327412287183
x62=72.2564974285085x_{62} = 72.2564974285085
x63=53.4072061236969x_{63} = 53.4072061236969
x64=21.9912781084223x_{64} = 21.9912781084223
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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