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Plotting a function graph/ 2*cos(2*x)

Graphing y = 2*cos(2*x)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = 2*cos(2*x)
f(x)=2cos⁡(2x)f{\left(x \right)} = 2 \cos{\left(2 x \right)} 
f = 2*cos(2*x)
The graph of the function
0.000.250.500.751.001.251.501.752.002.252.502.753.005-5
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:
2cos⁡(2x)=02 \cos{\left(2 x \right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=π4x_{1} = \frac{\pi}{4}
x2=3π4x_{2} = \frac{3 \pi}{4}
Numerical solution
x1=−41.6261026600648x_{1} = -41.6261026600648
x2=162.577419823272x_{2} = 162.577419823272
x3=18.0641577581413x_{3} = 18.0641577581413
x4=25.9181393921158x_{4} = 25.9181393921158
x5=41.6261026600648x_{5} = 41.6261026600648
x6=40.0553063332699x_{6} = 40.0553063332699
x7=63.6172512351933x_{7} = 63.6172512351933
x8=−47.9092879672443x_{8} = -47.9092879672443
x9=88.7499924639117x_{9} = 88.7499924639117
x10=96.6039740978861x_{10} = 96.6039740978861
x11=82.4668071567321x_{11} = 82.4668071567321
x12=−13.3517687777566x_{12} = -13.3517687777566
x13=−62.0464549083984x_{13} = -62.0464549083984
x14=44.7676953136546x_{14} = 44.7676953136546
x15=−24.3473430653209x_{15} = -24.3473430653209
x16=−16.4933614313464x_{16} = -16.4933614313464
x17=19.6349540849362x_{17} = 19.6349540849362
x18=−3.92699081698724x_{18} = -3.92699081698724
x19=52.621676947629x_{19} = 52.621676947629
x20=−84.037603483527x_{20} = -84.037603483527
x21=−49.4800842940392x_{21} = -49.4800842940392
x22=−63.6172512351933x_{22} = -63.6172512351933
x23=55.7632696012188x_{23} = 55.7632696012188
x24=91.8915851175014x_{24} = 91.8915851175014
x25=60.4756585816035x_{25} = 60.4756585816035
x26=−10.2101761241668x_{26} = -10.2101761241668
x27=32.2013246992954x_{27} = 32.2013246992954
x28=−85.6083998103219x_{28} = -85.6083998103219
x29=33.7721210260903x_{29} = 33.7721210260903
x30=−57.3340659280137x_{30} = -57.3340659280137
x31=−32.2013246992954x_{31} = -32.2013246992954
x32=30.6305283725005x_{32} = 30.6305283725005
x33=−82.4668071567321x_{33} = -82.4668071567321
x34=3.92699081698724x_{34} = 3.92699081698724
x35=2.35619449019234x_{35} = 2.35619449019234
x36=16.4933614313464x_{36} = 16.4933614313464
x37=76.1836218495525x_{37} = 76.1836218495525
x38=−2.35619449019234x_{38} = -2.35619449019234
x39=−90.3207887907066x_{39} = -90.3207887907066
x40=−46.3384916404494x_{40} = -46.3384916404494
x41=−98.174770424681x_{41} = -98.174770424681
x42=46.3384916404494x_{42} = 46.3384916404494
x43=5.49778714378214x_{43} = 5.49778714378214
x44=47.9092879672443x_{44} = 47.9092879672443
x45=99.7455667514759x_{45} = 99.7455667514759
x46=−91.8915851175014x_{46} = -91.8915851175014
x47=1973.70558461779x_{47} = 1973.70558461779
x48=−33.7721210260903x_{48} = -33.7721210260903
x49=−19.6349540849362x_{49} = -19.6349540849362
x50=−60.4756585816035x_{50} = -60.4756585816035
x51=27.4889357189107x_{51} = 27.4889357189107
x52=−71.4712328691678x_{52} = -71.4712328691678
x53=84.037603483527x_{53} = 84.037603483527
x54=−76.1836218495525x_{54} = -76.1836218495525
x55=−12461.9126586273x_{55} = -12461.9126586273
x56=−38.484510006475x_{56} = -38.484510006475
x57=22.776546738526x_{57} = 22.776546738526
x58=68.329640215578x_{58} = 68.329640215578
x59=66.7588438887831x_{59} = 66.7588438887831
x60=−11.7809724509617x_{60} = -11.7809724509617
x61=−27.4889357189107x_{61} = -27.4889357189107
x62=62.0464549083984x_{62} = 62.0464549083984
x63=−25.9181393921158x_{63} = -25.9181393921158
x64=−40.0553063332699x_{64} = -40.0553063332699
x65=−5.49778714378214x_{65} = -5.49778714378214
x66=54.1924732744239x_{66} = 54.1924732744239
x67=−35.3429173528852x_{67} = -35.3429173528852
x68=384.059701901352x_{68} = 384.059701901352
x69=90.3207887907066x_{69} = 90.3207887907066
x70=−55.7632696012188x_{70} = -55.7632696012188
x71=11.7809724509617x_{71} = 11.7809724509617
x72=8.63937979737193x_{72} = 8.63937979737193
x73=74.6128255227576x_{73} = 74.6128255227576
x74=−54.1924732744239x_{74} = -54.1924732744239
x75=−99.7455667514759x_{75} = -99.7455667514759
x76=−18.0641577581413x_{76} = -18.0641577581413
x77=38.484510006475x_{77} = 38.484510006475
x78=−77.7544181763474x_{78} = -77.7544181763474
x79=−79.3252145031423x_{79} = -79.3252145031423
x80=10.2101761241668x_{80} = 10.2101761241668
x81=85.6083998103219x_{81} = 85.6083998103219
x82=98.174770424681x_{82} = 98.174770424681
x83=−68.329640215578x_{83} = -68.329640215578
x84=−69.9004365423729x_{84} = -69.9004365423729
x85=49.4800842940392x_{85} = 49.4800842940392
x86=69.9004365423729x_{86} = 69.9004365423729
x87=24.3473430653209x_{87} = 24.3473430653209
x88=−93.4623814442964x_{88} = -93.4623814442964
x89=77.7544181763474x_{89} = 77.7544181763474
x90=87.1791961371168x_{90} = 87.1791961371168
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 2*cos(2*x).
2cos⁡(0⋅2)2 \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
−4sin⁡(2x)=0- 4 \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     
(--, -2)
 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
−8cos⁡(2x)=0- 8 \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→−∞(2cos⁡(2x))=⟨−2,2⟩\lim_{x \to -\infty}\left(2 \cos{\left(2 x \right)}\right) = \left\langle -2, 2\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=⟨−2,2⟩y = \left\langle -2, 2\right\rangle
lim⁡x→∞(2cos⁡(2x))=⟨−2,2⟩\lim_{x \to \infty}\left(2 \cos{\left(2 x \right)}\right) = \left\langle -2, 2\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=⟨−2,2⟩y = \left\langle -2, 2\right\rangle
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 2*cos(2*x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(2cos⁡(2x)x)=0\lim_{x \to -\infty}\left(\frac{2 \cos{\left(2 x \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(2cos⁡(2x)x)=0\lim_{x \to \infty}\left(\frac{2 \cos{\left(2 x \right)}}{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:
2cos⁡(2x)=2cos⁡(2x)2 \cos{\left(2 x \right)} = 2 \cos{\left(2 x \right)}
- Yes
2cos⁡(2x)=−2cos⁡(2x)2 \cos{\left(2 x \right)} = - 2 \cos{\left(2 x \right)}
- No
so, the function
is
even
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