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

Graphing y = sqrt(2)*cos(x)

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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)} = \sqrt{2} \cos{\left(x \right)} 
f = sqrt(2)*cos(x)
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)=0\sqrt{2} \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=48.6946861306418x_{1} = 48.6946861306418
x2=54.9778714378214x_{2} = 54.9778714378214
x3=−98.9601685880785x_{3} = -98.9601685880785
x4=67.5442420521806x_{4} = 67.5442420521806
x5=76.9690200129499x_{5} = 76.9690200129499
x6=36.1283155162826x_{6} = 36.1283155162826
x7=58.1194640914112x_{7} = 58.1194640914112
x8=14.1371669411541x_{8} = 14.1371669411541
x9=−29.845130209103x_{9} = -29.845130209103
x10=61.261056745001x_{10} = 61.261056745001
x11=−36.1283155162826x_{11} = -36.1283155162826
x12=−4.71238898038469x_{12} = -4.71238898038469
x13=−39.2699081698724x_{13} = -39.2699081698724
x14=1.5707963267949x_{14} = 1.5707963267949
x15=−14.1371669411541x_{15} = -14.1371669411541
x16=−64.4026493985908x_{16} = -64.4026493985908
x17=−67.5442420521806x_{17} = -67.5442420521806
x18=92.6769832808989x_{18} = 92.6769832808989
x19=−51.8362787842316x_{19} = -51.8362787842316
x20=−86.3937979737193x_{20} = -86.3937979737193
x21=42.4115008234622x_{21} = 42.4115008234622
x22=−17.2787595947439x_{22} = -17.2787595947439
x23=−45.553093477052x_{23} = -45.553093477052
x24=−472.809694365264x_{24} = -472.809694365264
x25=−89.5353906273091x_{25} = -89.5353906273091
x26=−1.5707963267949x_{26} = -1.5707963267949
x27=39.2699081698724x_{27} = 39.2699081698724
x28=23.5619449019235x_{28} = 23.5619449019235
x29=7.85398163397448x_{29} = 7.85398163397448
x30=−58.1194640914112x_{30} = -58.1194640914112
x31=−61.261056745001x_{31} = -61.261056745001
x32=−73.8274273593601x_{32} = -73.8274273593601
x33=73.8274273593601x_{33} = 73.8274273593601
x34=29.845130209103x_{34} = 29.845130209103
x35=4.71238898038469x_{35} = 4.71238898038469
x36=86.3937979737193x_{36} = 86.3937979737193
x37=64.4026493985908x_{37} = 64.4026493985908
x38=89.5353906273091x_{38} = 89.5353906273091
x39=−20.4203522483337x_{39} = -20.4203522483337
x40=−387.986692718339x_{40} = -387.986692718339
x41=−26.7035375555132x_{41} = -26.7035375555132
x42=98.9601685880785x_{42} = 98.9601685880785
x43=51.8362787842316x_{43} = 51.8362787842316
x44=83.2522053201295x_{44} = 83.2522053201295
x45=−48.6946861306418x_{45} = -48.6946861306418
x46=−54.9778714378214x_{46} = -54.9778714378214
x47=70.6858347057703x_{47} = 70.6858347057703
x48=−95.8185759344887x_{48} = -95.8185759344887
x49=26.7035375555132x_{49} = 26.7035375555132
x50=80.1106126665397x_{50} = 80.1106126665397
x51=−23.5619449019235x_{51} = -23.5619449019235
x52=−7.85398163397448x_{52} = -7.85398163397448
x53=−83.2522053201295x_{53} = -83.2522053201295
x54=−76.9690200129499x_{54} = -76.9690200129499
x55=−42.4115008234622x_{55} = -42.4115008234622
x56=−32.9867228626928x_{56} = -32.9867228626928
x57=−3626.96871856942x_{57} = -3626.96871856942
x58=17.2787595947439x_{58} = 17.2787595947439
x59=32.9867228626928x_{59} = 32.9867228626928
x60=20.4203522483337x_{60} = 20.4203522483337
x61=−70.6858347057703x_{61} = -70.6858347057703
x62=−10.9955742875643x_{62} = -10.9955742875643
x63=−92.6769832808989x_{63} = -92.6769832808989
x64=45.553093477052x_{64} = 45.553093477052
x65=10.9955742875643x_{65} = 10.9955742875643
x66=−80.1106126665397x_{66} = -80.1106126665397
x67=95.8185759344887x_{67} = 95.8185759344887
x68=−2266.65909956504x_{68} = -2266.65909956504
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to sqrt(2)*cos(x).
2cos⁡(0)\sqrt{2} \cos{\left(0 \right)}
The result:
f(0)=2f{\left(0 \right)} = \sqrt{2}
The point:
(0, sqrt(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- \sqrt{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- \sqrt{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⟨−1,1⟩\lim_{x \to -\infty}\left(\sqrt{2} \cos{\left(x \right)}\right) = \sqrt{2} \left\langle -1, 1\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=2⟨−1,1⟩y = \sqrt{2} \left\langle -1, 1\right\rangle
lim⁡x→∞(2cos⁡(x))=2⟨−1,1⟩\lim_{x \to \infty}\left(\sqrt{2} \cos{\left(x \right)}\right) = \sqrt{2} \left\langle -1, 1\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=2⟨−1,1⟩y = \sqrt{2} \left\langle -1, 1\right\rangle
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of sqrt(2)*cos(x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(2cos⁡(x)x)=0\lim_{x \to -\infty}\left(\frac{\sqrt{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{\sqrt{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)\sqrt{2} \cos{\left(x \right)} = \sqrt{2} \cos{\left(x \right)}
- Yes
2cos⁡(x)=−2cos⁡(x)\sqrt{2} \cos{\left(x \right)} = - \sqrt{2} \cos{\left(x \right)}
- No
so, the function
is
even
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