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Plotting a function graph/ cos(x)-1/2

Graphing y = cos(x)-1/2

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = cos(x) - 1/2
f(x)=cos⁡(x)−12f{\left(x \right)} = \cos{\left(x \right)} - \frac{1}{2} 
f = cos(x) - 1/2
The graph of the function
02468-8-6-4-2-10102-2
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⁡(x)−12=0\cos{\left(x \right)} - \frac{1}{2} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=π3x_{1} = \frac{\pi}{3}
x2=5π3x_{2} = \frac{5 \pi}{3}
Numerical solution
x1=−99.4837673636768x_{1} = -99.4837673636768
x2=−24.0855436775217x_{2} = -24.0855436775217
x3=−36.6519142918809x_{3} = -36.6519142918809
x4=−19.8967534727354x_{4} = -19.8967534727354
x5=55.5014702134197x_{5} = 55.5014702134197
x6=38.7463093942741x_{6} = 38.7463093942741
x7=−74.3510261349584x_{7} = -74.3510261349584
x8=76.4454212373516x_{8} = 76.4454212373516
x9=42.9350995990605x_{9} = 42.9350995990605
x10=−89.0117918517108x_{10} = -89.0117918517108
x11=68.0678408277789x_{11} = 68.0678408277789
x12=49.2182849062401x_{12} = 49.2182849062401
x13=−86.9173967493176x_{13} = -86.9173967493176
x14=24.0855436775217x_{14} = 24.0855436775217
x15=−70.162235930172x_{15} = -70.162235930172
x16=−26.1799387799149x_{16} = -26.1799387799149
x17=5.23598775598299x_{17} = 5.23598775598299
x18=−1.0471975511966x_{18} = -1.0471975511966
x19=17.8023583703422x_{19} = 17.8023583703422
x20=−95.2949771588904x_{20} = -95.2949771588904
x21=−38.7463093942741x_{21} = -38.7463093942741
x22=−225.147473507269x_{22} = -225.147473507269
x23=95.2949771588904x_{23} = 95.2949771588904
x24=93.2005820564972x_{24} = 93.2005820564972
x25=−5.23598775598299x_{25} = -5.23598775598299
x26=19.8967534727354x_{26} = 19.8967534727354
x27=74.3510261349584x_{27} = 74.3510261349584
x28=99.4837673636768x_{28} = 99.4837673636768
x29=45.0294947014537x_{29} = 45.0294947014537
x30=86.9173967493176x_{30} = 86.9173967493176
x31=63.8790506229925x_{31} = 63.8790506229925
x32=82.7286065445312x_{32} = 82.7286065445312
x33=80.634211442138x_{33} = 80.634211442138
x34=−7.33038285837618x_{34} = -7.33038285837618
x35=−42.9350995990605x_{35} = -42.9350995990605
x36=51.3126800086333x_{36} = 51.3126800086333
x37=26.1799387799149x_{37} = 26.1799387799149
x38=11.5191730631626x_{38} = 11.5191730631626
x39=−93.2005820564972x_{39} = -93.2005820564972
x40=−45.0294947014537x_{40} = -45.0294947014537
x41=36.6519142918809x_{41} = 36.6519142918809
x42=7.33038285837618x_{42} = 7.33038285837618
x43=89.0117918517108x_{43} = 89.0117918517108
x44=32.4631240870945x_{44} = 32.4631240870945
x45=70.162235930172x_{45} = 70.162235930172
x46=−82.7286065445312x_{46} = -82.7286065445312
x47=−63.8790506229925x_{47} = -63.8790506229925
x48=−13.6135681655558x_{48} = -13.6135681655558
x49=−61.7846555205993x_{49} = -61.7846555205993
x50=30.3687289847013x_{50} = 30.3687289847013
x51=−359.188760060433x_{51} = -359.188760060433
x52=−11.5191730631626x_{52} = -11.5191730631626
x53=−68.0678408277789x_{53} = -68.0678408277789
x54=−30.3687289847013x_{54} = -30.3687289847013
x55=−76.4454212373516x_{55} = -76.4454212373516
x56=1651.43053823704x_{56} = 1651.43053823704
x57=57.5958653158129x_{57} = 57.5958653158129
x58=−17.8023583703422x_{58} = -17.8023583703422
x59=1.0471975511966x_{59} = 1.0471975511966
x60=−55.5014702134197x_{60} = -55.5014702134197
x61=−80.634211442138x_{61} = -80.634211442138
x62=13.6135681655558x_{62} = 13.6135681655558
x63=−51.3126800086333x_{63} = -51.3126800086333
x64=−49.2182849062401x_{64} = -49.2182849062401
x65=−32.4631240870945x_{65} = -32.4631240870945
x66=−57.5958653158129x_{66} = -57.5958653158129
x67=61.7846555205993x_{67} = 61.7846555205993
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to cos(x) - 1/2.
−12+cos⁡(0)- \frac{1}{2} + \cos{\left(0 \right)}
The result:
f(0)=12f{\left(0 \right)} = \frac{1}{2}
The point:
(0, 1/2)
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
−sin⁡(x)=0- \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, 1/2)

(pi, -3/2)


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
−cos⁡(x)=0- \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→−∞(cos⁡(x)−12)=⟨−32,12⟩\lim_{x \to -\infty}\left(\cos{\left(x \right)} - \frac{1}{2}\right) = \left\langle - \frac{3}{2}, \frac{1}{2}\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=⟨−32,12⟩y = \left\langle - \frac{3}{2}, \frac{1}{2}\right\rangle
lim⁡x→∞(cos⁡(x)−12)=⟨−32,12⟩\lim_{x \to \infty}\left(\cos{\left(x \right)} - \frac{1}{2}\right) = \left\langle - \frac{3}{2}, \frac{1}{2}\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=⟨−32,12⟩y = \left\langle - \frac{3}{2}, \frac{1}{2}\right\rangle
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of cos(x) - 1/2, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(cos⁡(x)−12x)=0\lim_{x \to -\infty}\left(\frac{\cos{\left(x \right)} - \frac{1}{2}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(cos⁡(x)−12x)=0\lim_{x \to \infty}\left(\frac{\cos{\left(x \right)} - \frac{1}{2}}{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⁡(x)−12=cos⁡(x)−12\cos{\left(x \right)} - \frac{1}{2} = \cos{\left(x \right)} - \frac{1}{2}
- Yes
cos⁡(x)−12=12−cos⁡(x)\cos{\left(x \right)} - \frac{1}{2} = \frac{1}{2} - \cos{\left(x \right)}
- No
so, the function
is
even
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