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2cosx
Plotting a function graph/ 2cosx

Graphing y = 2cosx

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = 2*cos(x)
f(x)=2cos⁡(x)f{\left(x \right)} = 2 \cos{\left(x \right)} 
f = 2*cos(x)
The graph of the function
0-30-20-101020304050607080905-5
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⁡(x)=02 \cos{\left(x \right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=π2x_{1} = \frac{\pi}{2}
x2=3π2x_{2} = \frac{3 \pi}{2}
Numerical solution
x1=36.1283155162826x_{1} = 36.1283155162826
x2=−36.1283155162826x_{2} = -36.1283155162826
x3=98.9601685880785x_{3} = 98.9601685880785
x4=17.2787595947439x_{4} = 17.2787595947439
x5=−48.6946861306418x_{5} = -48.6946861306418
x6=−10.9955742875643x_{6} = -10.9955742875643
x7=−387.986692718339x_{7} = -387.986692718339
x8=48.6946861306418x_{8} = 48.6946861306418
x9=−92.6769832808989x_{9} = -92.6769832808989
x10=45.553093477052x_{10} = 45.553093477052
x11=−17.2787595947439x_{11} = -17.2787595947439
x12=61.261056745001x_{12} = 61.261056745001
x13=−168.075206967054x_{13} = -168.075206967054
x14=23.5619449019235x_{14} = 23.5619449019235
x15=76.9690200129499x_{15} = 76.9690200129499
x16=−86.3937979737193x_{16} = -86.3937979737193
x17=−1.5707963267949x_{17} = -1.5707963267949
x18=−4.71238898038469x_{18} = -4.71238898038469
x19=73.8274273593601x_{19} = 73.8274273593601
x20=51.8362787842316x_{20} = 51.8362787842316
x21=−70.6858347057703x_{21} = -70.6858347057703
x22=−58.1194640914112x_{22} = -58.1194640914112
x23=−54.9778714378214x_{23} = -54.9778714378214
x24=−98.9601685880785x_{24} = -98.9601685880785
x25=−61.261056745001x_{25} = -61.261056745001
x26=70.6858347057703x_{26} = 70.6858347057703
x27=58.1194640914112x_{27} = 58.1194640914112
x28=−23.5619449019235x_{28} = -23.5619449019235
x29=−80.1106126665397x_{29} = -80.1106126665397
x30=83.2522053201295x_{30} = 83.2522053201295
x31=1.5707963267949x_{31} = 1.5707963267949
x32=67.5442420521806x_{32} = 67.5442420521806
x33=−64.4026493985908x_{33} = -64.4026493985908
x34=−51.8362787842316x_{34} = -51.8362787842316
x35=7.85398163397448x_{35} = 7.85398163397448
x36=−26.7035375555132x_{36} = -26.7035375555132
x37=−45.553093477052x_{37} = -45.553093477052
x38=86.3937979737193x_{38} = 86.3937979737193
x39=10.9955742875643x_{39} = 10.9955742875643
x40=−73.8274273593601x_{40} = -73.8274273593601
x41=−7.85398163397448x_{41} = -7.85398163397448
x42=92.6769832808989x_{42} = 92.6769832808989
x43=80.1106126665397x_{43} = 80.1106126665397
x44=4.71238898038469x_{44} = 4.71238898038469
x45=32.9867228626928x_{45} = 32.9867228626928
x46=20.4203522483337x_{46} = 20.4203522483337
x47=−95.8185759344887x_{47} = -95.8185759344887
x48=−83.2522053201295x_{48} = -83.2522053201295
x49=−89.5353906273091x_{49} = -89.5353906273091
x50=64.4026493985908x_{50} = 64.4026493985908
x51=−39.2699081698724x_{51} = -39.2699081698724
x52=−20.4203522483337x_{52} = -20.4203522483337
x53=−67.5442420521806x_{53} = -67.5442420521806
x54=42.4115008234622x_{54} = 42.4115008234622
x55=26.7035375555132x_{55} = 26.7035375555132
x56=−14.1371669411541x_{56} = -14.1371669411541
x57=−76.9690200129499x_{57} = -76.9690200129499
x58=89.5353906273091x_{58} = 89.5353906273091
x59=95.8185759344887x_{59} = 95.8185759344887
x60=14.1371669411541x_{60} = 14.1371669411541
x61=−42.4115008234622x_{61} = -42.4115008234622
x62=−2266.65909956504x_{62} = -2266.65909956504
x63=54.9778714378214x_{63} = 54.9778714378214
x64=−32.9867228626928x_{64} = -32.9867228626928
x65=−29.845130209103x_{65} = -29.845130209103
x66=39.2699081698724x_{66} = 39.2699081698724
x67=29.845130209103x_{67} = 29.845130209103
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 2*cos(x).
2cos⁡(0)2 \cos{\left(0 \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⁡(x)=0- 2 \sin{\left(x \right)} = 0
Solve this equation
The roots of this equation
x1=0x_{1} = 0
x2=πx_{2} = \pi
The values of the extrema at the points:
(0, 2)

(pi, -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=πx_{1} = \pi
Maxima of the function at points:
x1=0x_{1} = 0
Decreasing at intervals
(−∞,0]∪[π,∞)\left(-\infty, 0\right] \cup \left[\pi, \infty\right)
Increasing at intervals
[0,π]\left[0, \pi\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
−2cos⁡(x)=0- 2 \cos{\left(x \right)} = 0
Solve this equation
The roots of this equation
x1=π2x_{1} = \frac{\pi}{2}
x2=3π2x_{2} = \frac{3 \pi}{2}

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
[π2,3π2]\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]
Convex at the intervals
(−∞,π2]∪[3π2,∞)\left(-\infty, \frac{\pi}{2}\right] \cup \left[\frac{3 \pi}{2}, \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⁡(x))=⟨−2,2⟩\lim_{x \to -\infty}\left(2 \cos{\left(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⁡(x))=⟨−2,2⟩\lim_{x \to \infty}\left(2 \cos{\left(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(x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(2cos⁡(x)x)=0\lim_{x \to -\infty}\left(\frac{2 \cos{\left(x \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(2cos⁡(x)x)=0\lim_{x \to \infty}\left(\frac{2 \cos{\left(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⁡(x)=2cos⁡(x)2 \cos{\left(x \right)} = 2 \cos{\left(x \right)}
- Yes
2cos⁡(x)=−2cos⁡(x)2 \cos{\left(x \right)} = - 2 \cos{\left(x \right)}
- No
so, the function
is
even
The graph
Graphing y = 2cosx
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