Mister Exam

Other calculators

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

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

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
                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
0.00.51.01.52.02.53.03.54.04.55.05.56.02.5-2.5
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=−1083.8495084391x_{1} = -1083.8495084391
x2=62.8318530717959x_{2} = 62.8318530717959
x3=−50.2654824574367x_{3} = -50.2654824574367
x4=−3.14171741723949x_{4} = -3.14171741723949
x5=72.2566292957295x_{5} = 72.2566292957295
x6=−9.42485173622935x_{6} = -9.42485173622935
x7=−47.1240173901594x_{7} = -47.1240173901594
x8=78.5397496778866x_{8} = 78.5397496778866
x9=21.9911516419074x_{9} = 21.9911516419074
x10=−72.2565620594227x_{10} = -72.2565620594227
x11=15.7080397066029x_{11} = 15.7080397066029
x12=59.6903404916682x_{12} = 59.6903404916682
x13=72.2566368440238x_{13} = 72.2566368440238
x14=−97.3894529845737x_{14} = -97.3894529845737
x15=−6.28318530717959x_{15} = -6.28318530717959
x16=97.389506033414x_{16} = 97.389506033414
x17=72.2564974285085x_{17} = 72.2564974285085
x18=28.2743275355147x_{18} = 28.2743275355147
x19=53.4072061236969x_{19} = 53.4072061236969
x20=−75.398223686155x_{20} = -75.398223686155
x21=−53.4071523808127x_{21} = -53.4071523808127
x22=−43.9822971502571x_{22} = -43.9822971502571
x23=31.4159265358979x_{23} = 31.4159265358979
x24=−59.6902836920888x_{24} = -59.6902836920888
x25=28.2742362262029x_{25} = 28.2742362262029
x26=−69.1150383789755x_{26} = -69.1150383789755
x27=12.5663706143592x_{27} = 12.5663706143592
x28=87.9645943005142x_{28} = 87.9645943005142
x29=−28.2742611423571x_{29} = -28.2742611423571
x30=−37.6991118430775x_{30} = -37.6991118430775
x31=−78.5396939083992x_{31} = -78.5396939083992
x32=−91.1063173161218x_{32} = -91.1063173161218
x33=−100.530964914873x_{33} = -100.530964914873
x34=0x_{34} = 0
x35=−12.5663706143592x_{35} = -12.5663706143592
x36=9.42490616313103x_{36} = 9.42490616313103
x37=18.8495559215388x_{37} = 18.8495559215388
x38=−94.2477796076938x_{38} = -94.2477796076938
x39=−15.7080226365019x_{39} = -15.7080226365019
x40=43.9822971502571x_{40} = 43.9822971502571
x41=−31.4159265358979x_{41} = -31.4159265358979
x42=−15.7079741551755x_{42} = -15.7079741551755
x43=−81.6814089933346x_{43} = -81.6814089933346
x44=56.5486677646163x_{44} = 56.5486677646163
x45=−21.9911516405744x_{45} = -21.9911516405744
x46=40.8405826017537x_{46} = 40.8405826017537
x47=−56.5486677646163x_{47} = -56.5486677646163
x48=50.2654824574367x_{48} = 50.2654824574367
x49=−59.6902757594272x_{49} = -59.6902757594272
x50=94.2477796076938x_{50} = 94.2477796076938
x51=−65.9734547074718x_{51} = -65.9734547074718
x52=84.8228826659845x_{52} = 84.8228826659845
x53=65.9735385828884x_{53} = 65.9735385828884
x54=37.6991118430775x_{54} = 37.6991118430775
x55=−87.9645943005142x_{55} = -87.9645943005142
x56=34.557448949744x_{56} = 34.557448949744
x57=65.9734548161256x_{57} = 65.9734548161256
x58=−25.1327412287183x_{58} = -25.1327412287183
x59=81.6814089933346x_{59} = 81.6814089933346
x60=6.28318530717959x_{60} = 6.28318530717959
x61=−34.5573938477265x_{61} = -34.5573938477265
x62=100.530964914873x_{62} = 100.530964914873
x63=21.9912781084223x_{63} = 21.9912781084223
x64=75.398223686155x_{64} = 75.398223686155
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
    © Mister Exam – Calculator