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

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

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = 2*sin (x) + 2*sin(2*x) + 1
f(x)=(2sin⁡2(x)+2sin⁡(2x))+1f{\left(x \right)} = \left(2 \sin^{2}{\left(x \right)} + 2 \sin{\left(2 x \right)}\right) + 1 
f = 2*sin(x)^2 + 2*sin(2*x) + 1
The graph of the function
02468-8-6-4-2-10105-5
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:
(2sin⁡2(x)+2sin⁡(2x))+1=0\left(2 \sin^{2}{\left(x \right)} + 2 \sin{\left(2 x \right)}\right) + 1 = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−π4x_{1} = - \frac{\pi}{4}
x2=−ilog⁡((−i)52)x_{2} = - i \log{\left(\left(- i\right)^{\frac{5}{2}} \right)}
Numerical solution
x1=−79.3252145031423x_{1} = -79.3252145031423
x2=−85.6083998103219x_{2} = -85.6083998103219
x3=−50.5872330118333x_{3} = -50.5872330118333
x4=84.037603483527x_{4} = 84.037603483527
x5=−1608.81718919237x_{5} = -1608.81718919237
x6=24.3473430653209x_{6} = 24.3473430653209
x7=−6.60493586157623x_{7} = -6.60493586157623
x8=−7.06858347057703x_{8} = -7.06858347057703
x9=52.621676947629x_{9} = 52.621676947629
x10=46.3384916404494x_{10} = 46.3384916404494
x11=36.9137136796801x_{11} = 36.9137136796801
x12=−88.2863448549109x_{12} = -88.2863448549109
x13=−25.9181393921158x_{13} = -25.9181393921158
x14=−31.7376770902946x_{14} = -31.7376770902946
x15=−94.5695301620904x_{15} = -94.5695301620904
x16=−19.6349540849362x_{16} = -19.6349540849362
x17=21.6693980207319x_{17} = 21.6693980207319
x18=68.329640215578x_{18} = 68.329640215578
x19=−60.0120109726027x_{19} = -60.0120109726027
x20=96.6039740978861x_{20} = 96.6039740978861
x21=−91.8915851175014x_{21} = -91.8915851175014
x22=−13.3517687777566x_{22} = -13.3517687777566
x23=−44.3040477046537x_{23} = -44.3040477046537
x24=−73.0420291959627x_{24} = -73.0420291959627
x25=−72.5783815869619x_{25} = -72.5783815869619
x26=−66.2951962797823x_{26} = -66.2951962797823
x27=78.2180657853482x_{27} = 78.2180657853482
x28=27.9525833279115x_{28} = 27.9525833279115
x29=12.2446200599625x_{29} = 12.2446200599625
x30=−29.0597320457056x_{30} = -29.0597320457056
x31=−47.9092879672443x_{31} = -47.9092879672443
x32=100.209214360477x_{32} = 100.209214360477
x33=−95.0331777710912x_{33} = -95.0331777710912
x34=80.8960108299372x_{34} = 80.8960108299372
x35=81.359658438938x_{35} = 81.359658438938
x36=−638.528706842126x_{36} = -638.528706842126
x37=34.2357686350911x_{37} = 34.2357686350911
x38=−57.3340659280137x_{38} = -57.3340659280137
x39=8.63937979737193x_{39} = 8.63937979737193
x40=58.9048622548086x_{40} = 58.9048622548086
x41=40.0553063332699x_{41} = 40.0553063332699
x42=−69.9004365423729x_{42} = -69.9004365423729
x43=37.3773612886809x_{43} = 37.3773612886809
x44=87.6428437461176x_{44} = 87.6428437461176
x45=59.3685098638094x_{45} = 59.3685098638094
x46=90.3207887907066x_{46} = 90.3207887907066
x47=−69.4367889333721x_{47} = -69.4367889333721
x48=74.6128255227576x_{48} = 74.6128255227576
x49=−16.0297138223456x_{49} = -16.0297138223456
x50=5.96143475278294x_{50} = 5.96143475278294
x51=62.0464549083984x_{51} = 62.0464549083984
x52=43.6605465958605x_{52} = 43.6605465958605
x53=−85.1447522013211x_{53} = -85.1447522013211
x54=15.3862127135523x_{54} = 15.3862127135523
x55=−0.321750554396642x_{55} = -0.321750554396642
x56=−3.92699081698724x_{56} = -3.92699081698724
x57=−75.7199742405517x_{57} = -75.7199742405517
x58=−22.3128991295252x_{58} = -22.3128991295252
x59=−28.5960844367048x_{59} = -28.5960844367048
x60=−51.0508806208341x_{60} = -51.0508806208341
x61=−82.0031595477313x_{61} = -82.0031595477313
x62=65.651695170989x_{62} = 65.651695170989
x63=−97.7111228156802x_{63} = -97.7111228156802
x64=71.9348804781686x_{64} = 71.9348804781686
x65=−63.6172512351933x_{65} = -63.6172512351933
x66=93.9260290532972x_{66} = 93.9260290532972
x67=−38.0208623974742x_{67} = -38.0208623974742
x68=14.9225651045515x_{68} = 14.9225651045515
x69=2.35619449019234x_{69} = 2.35619449019234
x70=18.0641577581413x_{70} = 18.0641577581413
x71=49.9437319030401x_{71} = 49.9437319030401
x72=−9.74652851516602x_{72} = -9.74652851516602
x73=−41.6261026600648x_{73} = -41.6261026600648
x74=30.6305283725005x_{74} = 30.6305283725005
x75=−35.3429173528852x_{75} = -35.3429173528852
x76=−53.7288256654231x_{76} = -53.7288256654231
x77=56.2269172102196x_{77} = 56.2269172102196
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 2*sin(x)^2 + 2*sin(2*x) + 1.
(2sin⁡2(0)+2sin⁡(0⋅2))+1\left(2 \sin^{2}{\left(0 \right)} + 2 \sin{\left(0 \cdot 2 \right)}\right) + 1
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
4sin⁡(x)cos⁡(x)+4cos⁡(2x)=04 \sin{\left(x \right)} \cos{\left(x \right)} + 4 \cos{\left(2 x \right)} = 0
Solve this equation
Solutions are not found,
function may have no extrema
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(−sin⁡2(x)−2sin⁡(2x)+cos⁡2(x))=04 \left(- \sin^{2}{\left(x \right)} - 2 \sin{\left(2 x \right)} + \cos^{2}{\left(x \right)}\right) = 0
Solve this equation
The roots of this equation
x1=−ilog⁡(−eiatan⁡(43)4)x_{1} = - i \log{\left(- e^{\frac{i \operatorname{atan}{\left(\frac{4}{3} \right)}}{4}} \right)}
x2=atan⁡(43)4x_{2} = \frac{\operatorname{atan}{\left(\frac{4}{3} \right)}}{4}

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⁡(sin⁡(atan⁡(43)4)cos⁡(atan⁡(43)4))]\left(-\infty, - \pi + \operatorname{atan}{\left(\frac{\sin{\left(\frac{\operatorname{atan}{\left(\frac{4}{3} \right)}}{4} \right)}}{\cos{\left(\frac{\operatorname{atan}{\left(\frac{4}{3} \right)}}{4} \right)}} \right)}\right]
Convex at the intervals
[atan⁡(43)4,∞)\left[\frac{\operatorname{atan}{\left(\frac{4}{3} \right)}}{4}, \infty\right)
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞((2sin⁡2(x)+2sin⁡(2x))+1)=⟨−1,5⟩\lim_{x \to -\infty}\left(\left(2 \sin^{2}{\left(x \right)} + 2 \sin{\left(2 x \right)}\right) + 1\right) = \left\langle -1, 5\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=⟨−1,5⟩y = \left\langle -1, 5\right\rangle
lim⁡x→∞((2sin⁡2(x)+2sin⁡(2x))+1)=⟨−1,5⟩\lim_{x \to \infty}\left(\left(2 \sin^{2}{\left(x \right)} + 2 \sin{\left(2 x \right)}\right) + 1\right) = \left\langle -1, 5\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=⟨−1,5⟩y = \left\langle -1, 5\right\rangle
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 2*sin(x)^2 + 2*sin(2*x) + 1, divided by x at x->+oo and x ->-oo
lim⁡x→−∞((2sin⁡2(x)+2sin⁡(2x))+1x)=0\lim_{x \to -\infty}\left(\frac{\left(2 \sin^{2}{\left(x \right)} + 2 \sin{\left(2 x \right)}\right) + 1}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞((2sin⁡2(x)+2sin⁡(2x))+1x)=0\lim_{x \to \infty}\left(\frac{\left(2 \sin^{2}{\left(x \right)} + 2 \sin{\left(2 x \right)}\right) + 1}{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:
(2sin⁡2(x)+2sin⁡(2x))+1=2sin⁡2(x)−2sin⁡(2x)+1\left(2 \sin^{2}{\left(x \right)} + 2 \sin{\left(2 x \right)}\right) + 1 = 2 \sin^{2}{\left(x \right)} - 2 \sin{\left(2 x \right)} + 1
- No
(2sin⁡2(x)+2sin⁡(2x))+1=−2sin⁡2(x)+2sin⁡(2x)−1\left(2 \sin^{2}{\left(x \right)} + 2 \sin{\left(2 x \right)}\right) + 1 = - 2 \sin^{2}{\left(x \right)} + 2 \sin{\left(2 x \right)} - 1
- No
so, the function
not is
neither even, nor odd
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