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Plotting a function graph/ 1+cos2x

Graphing y = 1+cos2x

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = 1 + cos(2*x)
f(x)=cos⁡(2x)+1f{\left(x \right)} = \cos{\left(2 x \right)} + 1 
f = cos(2*x) + 1
The graph of the function
02468-8-6-4-2-101004
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:
cos⁡(2x)+1=0\cos{\left(2 x \right)} + 1 = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=π2x_{1} = \frac{\pi}{2}
Numerical solution
x1=10.9955743696636x_{1} = 10.9955743696636
x2=−7.85398149857354x_{2} = -7.85398149857354
x3=−1.57079642969308x_{3} = -1.57079642969308
x4=45.553093700501x_{4} = 45.553093700501
x5=29.845130320338x_{5} = 29.845130320338
x6=−73.8274272800405x_{6} = -73.8274272800405
x7=89.5353908552844x_{7} = 89.5353908552844
x8=80.1106131434937x_{8} = 80.1106131434937
x9=−89.5353907467661x_{9} = -89.5353907467661
x10=−14.1371668392726x_{10} = -14.1371668392726
x11=−92.6769830239371x_{11} = -92.6769830239371
x12=541.924732890135x_{12} = 541.924732890135
x13=48.6946859238715x_{13} = 48.6946859238715
x14=54.9778711883962x_{14} = 54.9778711883962
x15=80.1106126771746x_{15} = 80.1106126771746
x16=−32.9867227513827x_{16} = -32.9867227513827
x17=−45.5530935883361x_{17} = -45.5530935883361
x18=−76.9690202568697x_{18} = -76.9690202568697
x19=−4.71238872430683x_{19} = -4.71238872430683
x20=67.5442422779275x_{20} = 67.5442422779275
x21=92.6769830795146x_{21} = 92.6769830795146
x22=−61.2610562242523x_{22} = -61.2610562242523
x23=61.2610566752601x_{23} = 61.2610566752601
x24=−17.2787598091171x_{24} = -17.2787598091171
x25=76.9690207492347x_{25} = 76.9690207492347
x26=−54.9778716831146x_{26} = -54.9778716831146
x27=−4.7123889912442x_{27} = -4.7123889912442
x28=17.2787598502655x_{28} = 17.2787598502655
x29=64.4026493086922x_{29} = 64.4026493086922
x30=83.2522052340866x_{30} = 83.2522052340866
x31=20.4203521497111x_{31} = 20.4203521497111
x32=76.9690200400775x_{32} = 76.9690200400775
x33=83.2522055730903x_{33} = 83.2522055730903
x34=26.7035373461441x_{34} = 26.7035373461441
x35=−70.6858346386357x_{35} = -70.6858346386357
x36=−10.9955745350309x_{36} = -10.9955745350309
x37=39.2699081179815x_{37} = 39.2699081179815
x38=−48.6946860920117x_{38} = -48.6946860920117
x39=−29.8451300963672x_{39} = -29.8451300963672
x40=−42.4115006098842x_{40} = -42.4115006098842
x41=−39.2699083866483x_{41} = -39.2699083866483
x42=−86.393797765473x_{42} = -86.393797765473
x43=−76.9690198771149x_{43} = -76.9690198771149
x44=−83.2522055415057x_{44} = -83.2522055415057
x45=39.2699084246933x_{45} = 39.2699084246933
x46=−17.2787590276524x_{46} = -17.2787590276524
x47=86.393797888273x_{47} = 86.393797888273
x48=7.85398174058521x_{48} = 7.85398174058521
x49=−20.4203520321877x_{49} = -20.4203520321877
x50=70.6858345016621x_{50} = 70.6858345016621
x51=17.2787595624179x_{51} = 17.2787595624179
x52=−80.1106125795659x_{52} = -80.1106125795659
x53=−26.7035372990183x_{53} = -26.7035372990183
x54=73.8274274795554x_{54} = 73.8274274795554
x55=54.9778714849733x_{55} = 54.9778714849733
x56=98.9601683381274x_{56} = 98.9601683381274
x57=23.5619449395428x_{57} = 23.5619449395428
x58=−48.6946858738636x_{58} = -48.6946858738636
x59=−70.685834448838x_{59} = -70.685834448838
x60=23.5619451230057x_{60} = 23.5619451230057
x61=−67.5442421675773x_{61} = -67.5442421675773
x62=−39.2699081528781x_{62} = -39.2699081528781
x63=4.71238876848081x_{63} = 4.71238876848081
x64=−23.5619450090417x_{64} = -23.5619450090417
x65=−98.96016883042x_{65} = -98.96016883042
x66=−61.2610569641117x_{66} = -61.2610569641117
x67=14.1371671048484x_{67} = 14.1371671048484
x68=−64.4026491876462x_{68} = -64.4026491876462
x69=36.1283156002139x_{69} = 36.1283156002139
x70=−98.9601684414698x_{70} = -98.9601684414698
x71=−36.1283154192437x_{71} = -36.1283154192437
x72=98.9601685932308x_{72} = 98.9601685932308
x73=32.9867226137576x_{73} = 32.9867226137576
x74=−58.1194639993376x_{74} = -58.1194639993376
x75=1.5707965454425x_{75} = 1.5707965454425
x76=−54.9778713137198x_{76} = -54.9778713137198
x77=51.8362788999928x_{77} = 51.8362788999928
x78=42.4115007291722x_{78} = 42.4115007291722
x79=−26.7035375427973x_{79} = -26.7035375427973
x80=61.2610569989704x_{80} = 61.2610569989704
x81=−51.8362786897497x_{81} = -51.8362786897497
x82=−10.9955741902138x_{82} = -10.9955741902138
x83=10.9955740392793x_{83} = 10.9955740392793
x84=58.1194644379895x_{84} = 58.1194644379895
x85=−98.960168684456x_{85} = -98.960168684456
x86=76.9690197631883x_{86} = 76.9690197631883
x87=32.986722928111x_{87} = 32.986722928111
x88=−95.8185758681287x_{88} = -95.8185758681287
x89=−32.9867231091652x_{89} = -32.9867231091652
x90=−92.6769831823972x_{90} = -92.6769831823972
x91=95.8185760590309x_{91} = 95.8185760590309
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 1 + cos(2*x).
1+cos⁡(0⋅2)1 + \cos{\left(0 \cdot 2 \right)}
The result:
f(0)=2f{\left(0 \right)} = 2
The point:
(0, 2)
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
−2sin⁡(2x)=0- 2 \sin{\left(2 x \right)} = 0
Solve this equation
The roots of this equation
x1=0x_{1} = 0
x2=π2x_{2} = \frac{\pi}{2}
The values of the extrema at the points:
(0, 2)

 pi    
(--, 0)
 2


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=π2x_{1} = \frac{\pi}{2}
Maxima of the function at points:
x1=0x_{1} = 0
Decreasing at intervals
(−∞,0]∪[π2,∞)\left(-\infty, 0\right] \cup \left[\frac{\pi}{2}, \infty\right)
Increasing at intervals
[0,π2]\left[0, \frac{\pi}{2}\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
−4cos⁡(2x)=0- 4 \cos{\left(2 x \right)} = 0
Solve this equation
The roots of this equation
x1=π4x_{1} = \frac{\pi}{4}
x2=3π4x_{2} = \frac{3 \pi}{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
[π4,3π4]\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]
Convex at the intervals
(−∞,π4]∪[3π4,∞)\left(-\infty, \frac{\pi}{4}\right] \cup \left[\frac{3 \pi}{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→−∞(cos⁡(2x)+1)=⟨0,2⟩\lim_{x \to -\infty}\left(\cos{\left(2 x \right)} + 1\right) = \left\langle 0, 2\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=⟨0,2⟩y = \left\langle 0, 2\right\rangle
lim⁡x→∞(cos⁡(2x)+1)=⟨0,2⟩\lim_{x \to \infty}\left(\cos{\left(2 x \right)} + 1\right) = \left\langle 0, 2\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=⟨0,2⟩y = \left\langle 0, 2\right\rangle
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 1 + cos(2*x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(cos⁡(2x)+1x)=0\lim_{x \to -\infty}\left(\frac{\cos{\left(2 x \right)} + 1}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(cos⁡(2x)+1x)=0\lim_{x \to \infty}\left(\frac{\cos{\left(2 x \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:
cos⁡(2x)+1=cos⁡(2x)+1\cos{\left(2 x \right)} + 1 = \cos{\left(2 x \right)} + 1
- Yes
cos⁡(2x)+1=−cos⁡(2x)−1\cos{\left(2 x \right)} + 1 = - \cos{\left(2 x \right)} - 1
- No
so, the function
is
even
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