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Plotting a function graph/ 2*cos(x)+1

Graphing y = 2*cos(x)+1

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = 2*cos(x) + 1
f(x)=2cos⁡(x)+1f{\left(x \right)} = 2 \cos{\left(x \right)} + 1 
f = 2*cos(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:
2cos⁡(x)+1=02 \cos{\left(x \right)} + 1 = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=2π3x_{1} = \frac{2 \pi}{3}
x2=4π3x_{2} = \frac{4 \pi}{3}
Numerical solution
x1=90.0589894029074x_{1} = 90.0589894029074
x2=52.3598775598299x_{2} = 52.3598775598299
x3=39.7935069454707x_{3} = 39.7935069454707
x4=33.5103216382911x_{4} = 33.5103216382911
x5=−48.1710873550435x_{5} = -48.1710873550435
x6=−14.6607657167524x_{6} = -14.6607657167524
x7=−33.5103216382911x_{7} = -33.5103216382911
x8=−46.0766922526503x_{8} = -46.0766922526503
x9=−64.9262481741891x_{9} = -64.9262481741891
x10=−35.6047167406843x_{10} = -35.6047167406843
x11=8.37758040957278x_{11} = 8.37758040957278
x12=−52.3598775598299x_{12} = -52.3598775598299
x13=−16.7551608191456x_{13} = -16.7551608191456
x14=−27.2271363311115x_{14} = -27.2271363311115
x15=−90.0589894029074x_{15} = -90.0589894029074
x16=−8.37758040957278x_{16} = -8.37758040957278
x17=630.412925820352x_{17} = 630.412925820352
x18=−73.3038285837618x_{18} = -73.3038285837618
x19=−92.1533845053006x_{19} = -92.1533845053006
x20=416.784625376246x_{20} = 416.784625376246
x21=−96.342174710087x_{21} = -96.342174710087
x22=85.870199198121x_{22} = 85.870199198121
x23=−67.0206432765823x_{23} = -67.0206432765823
x24=27.2271363311115x_{24} = 27.2271363311115
x25=10.471975511966x_{25} = 10.471975511966
x26=−85.870199198121x_{26} = -85.870199198121
x27=−29.3215314335047x_{27} = -29.3215314335047
x28=60.7374579694027x_{28} = 60.7374579694027
x29=−54.4542726622231x_{29} = -54.4542726622231
x30=79.5870138909414x_{30} = 79.5870138909414
x31=58.6430628670095x_{31} = 58.6430628670095
x32=77.4926187885482x_{32} = 77.4926187885482
x33=64.9262481741891x_{33} = 64.9262481741891
x34=−77.4926187885482x_{34} = -77.4926187885482
x35=−39.7935069454707x_{35} = -39.7935069454707
x36=83.7758040957278x_{36} = 83.7758040957278
x37=117.286125734019x_{37} = 117.286125734019
x38=35.6047167406843x_{38} = 35.6047167406843
x39=−83.7758040957278x_{39} = -83.7758040957278
x40=41.8879020478639x_{40} = 41.8879020478639
x41=92.1533845053006x_{41} = 92.1533845053006
x42=−58.6430628670095x_{42} = -58.6430628670095
x43=−98.4365698124802x_{43} = -98.4365698124802
x44=−60.7374579694027x_{44} = -60.7374579694027
x45=73.3038285837618x_{45} = 73.3038285837618
x46=71.2094334813686x_{46} = 71.2094334813686
x47=48.1710873550435x_{47} = 48.1710873550435
x48=29.3215314335047x_{48} = 29.3215314335047
x49=−2.0943951023932x_{49} = -2.0943951023932
x50=20.943951023932x_{50} = 20.943951023932
x51=16.7551608191456x_{51} = 16.7551608191456
x52=−10.471975511966x_{52} = -10.471975511966
x53=54.4542726622231x_{53} = 54.4542726622231
x54=−79.5870138909414x_{54} = -79.5870138909414
x55=−1623.15620435473x_{55} = -1623.15620435473
x56=−23.0383461263252x_{56} = -23.0383461263252
x57=67.0206432765823x_{57} = 67.0206432765823
x58=98.4365698124802x_{58} = 98.4365698124802
x59=−41.8879020478639x_{59} = -41.8879020478639
x60=23.0383461263252x_{60} = 23.0383461263252
x61=−4.18879020478639x_{61} = -4.18879020478639
x62=2.0943951023932x_{62} = 2.0943951023932
x63=−20.943951023932x_{63} = -20.943951023932
x64=−71.2094334813686x_{64} = -71.2094334813686
x65=4.18879020478639x_{65} = 4.18879020478639
x66=46.0766922526503x_{66} = 46.0766922526503
x67=14.6607657167524x_{67} = 14.6607657167524
x68=96.342174710087x_{68} = 96.342174710087
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) + 1.
1+2cos⁡(0)1 + 2 \cos{\left(0 \right)}
The result:
f(0)=3f{\left(0 \right)} = 3
The point:
(0, 3)
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, 3)

(pi, -1)


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)+1)=⟨−1,3⟩\lim_{x \to -\infty}\left(2 \cos{\left(x \right)} + 1\right) = \left\langle -1, 3\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=⟨−1,3⟩y = \left\langle -1, 3\right\rangle
lim⁡x→∞(2cos⁡(x)+1)=⟨−1,3⟩\lim_{x \to \infty}\left(2 \cos{\left(x \right)} + 1\right) = \left\langle -1, 3\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=⟨−1,3⟩y = \left\langle -1, 3\right\rangle
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 2*cos(x) + 1, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(2cos⁡(x)+1x)=0\lim_{x \to -\infty}\left(\frac{2 \cos{\left(x \right)} + 1}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(2cos⁡(x)+1x)=0\lim_{x \to \infty}\left(\frac{2 \cos{\left(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:
2cos⁡(x)+1=2cos⁡(x)+12 \cos{\left(x \right)} + 1 = 2 \cos{\left(x \right)} + 1
- Yes
2cos⁡(x)+1=−2cos⁡(x)−12 \cos{\left(x \right)} + 1 = - 2 \cos{\left(x \right)} - 1
- No
so, the function
is
even
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