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Graphing y = 2cos5x+x^2

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = 2*cos(5*x) + x 
f(x)=x2+2cos(5x)f{\left(x \right)} = x^{2} + 2 \cos{\left(5 x \right)}
f = x^2 + 2*cos(5*x)
The graph of the function
02468-8-6-4-2-1010200-100
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:
x2+2cos(5x)=0x^{2} + 2 \cos{\left(5 x \right)} = 0
Solve this equation
The points of intersection with the axis X:

Numerical solution
x1=0.865665640694936x_{1} = 0.865665640694936
x2=0.324707661675468x_{2} = 0.324707661675468
x3=0.865665640694936x_{3} = -0.865665640694936
x4=0.324707661675468x_{4} = -0.324707661675468
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 2*cos(5*x) + x^2.
02+2cos(05)0^{2} + 2 \cos{\left(0 \cdot 5 \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
2x10sin(5x)=02 x - 10 \sin{\left(5 x \right)} = 0
Solve this equation
The roots of this equation
x1=2.62375881780545x_{1} = -2.62375881780545
x2=0.604095532292576x_{2} = -0.604095532292576
x3=3.95221773994643x_{3} = 3.95221773994643
x4=2.62375881780545x_{4} = 2.62375881780545
x5=0.604095532292576x_{5} = 0.604095532292576
x6=3.95221773994643x_{6} = -3.95221773994643
x7=4.19885728516055x_{7} = 4.19885728516055
x8=1.30964098411133x_{8} = -1.30964098411133
x9=3.01227783105993x_{9} = -3.01227783105993
x10=4.19885728516055x_{10} = -4.19885728516055
x11=1.8108371649809x_{11} = -1.8108371649809
x12=1.30964098411133x_{12} = 1.30964098411133
x13=0x_{13} = 0
x14=3.01227783105993x_{14} = 3.01227783105993
x15=1.8108371649809x_{15} = 1.8108371649809
The values of the extrema at the points:
(-2.6237588178054474, 8.5866209350238)

(-0.6040955322925761, -1.62041766902299)

(3.9522177399464318, 16.8450948537015)

(2.6237588178054474, 8.5866209350238)

(0.6040955322925761, -1.62041766902299)

(-3.9522177399464318, 16.8450948537015)

(4.198857285160547, 16.5445223973513)

(-1.3096409841113297, 3.64533423480503)

(-3.01227783105993, 7.47751288499468)

(-4.198857285160547, 16.5445223973513)

(-1.8108371649808959, 1.41490512466925)

(1.3096409841113297, 3.64533423480503)

(0, 2)

(3.01227783105993, 7.47751288499468)

(1.8108371649808959, 1.41490512466925)


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=0.604095532292576x_{1} = -0.604095532292576
x2=0.604095532292576x_{2} = 0.604095532292576
x3=4.19885728516055x_{3} = 4.19885728516055
x4=3.01227783105993x_{4} = -3.01227783105993
x5=4.19885728516055x_{5} = -4.19885728516055
x6=1.8108371649809x_{6} = -1.8108371649809
x7=3.01227783105993x_{7} = 3.01227783105993
x8=1.8108371649809x_{8} = 1.8108371649809
Maxima of the function at points:
x8=2.62375881780545x_{8} = -2.62375881780545
x8=3.95221773994643x_{8} = 3.95221773994643
x8=2.62375881780545x_{8} = 2.62375881780545
x8=3.95221773994643x_{8} = -3.95221773994643
x8=1.30964098411133x_{8} = -1.30964098411133
x8=1.30964098411133x_{8} = 1.30964098411133
x8=0x_{8} = 0
Decreasing at intervals
[4.19885728516055,)\left[4.19885728516055, \infty\right)
Increasing at intervals
(,4.19885728516055]\left(-\infty, -4.19885728516055\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
2(125cos(5x))=02 \left(1 - 25 \cos{\left(5 x \right)}\right) = 0
Solve this equation
The roots of this equation
x1=acos(125)5+2π5x_{1} = - \frac{\operatorname{acos}{\left(\frac{1}{25} \right)}}{5} + \frac{2 \pi}{5}
x2=acos(125)5x_{2} = \frac{\operatorname{acos}{\left(\frac{1}{25} \right)}}{5}

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
[acos(125)5,acos(125)5+2π5]\left[\frac{\operatorname{acos}{\left(\frac{1}{25} \right)}}{5}, - \frac{\operatorname{acos}{\left(\frac{1}{25} \right)}}{5} + \frac{2 \pi}{5}\right]
Convex at the intervals
(,acos(125)5][acos(125)5+2π5,)\left(-\infty, \frac{\operatorname{acos}{\left(\frac{1}{25} \right)}}{5}\right] \cup \left[- \frac{\operatorname{acos}{\left(\frac{1}{25} \right)}}{5} + \frac{2 \pi}{5}, \infty\right)
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
limx(x2+2cos(5x))=\lim_{x \to -\infty}\left(x^{2} + 2 \cos{\left(5 x \right)}\right) = \infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
limx(x2+2cos(5x))=\lim_{x \to \infty}\left(x^{2} + 2 \cos{\left(5 x \right)}\right) = \infty
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 2*cos(5*x) + x^2, divided by x at x->+oo and x ->-oo
limx(x2+2cos(5x)x)=\lim_{x \to -\infty}\left(\frac{x^{2} + 2 \cos{\left(5 x \right)}}{x}\right) = -\infty
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
limx(x2+2cos(5x)x)=\lim_{x \to \infty}\left(\frac{x^{2} + 2 \cos{\left(5 x \right)}}{x}\right) = \infty
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
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:
x2+2cos(5x)=x2+2cos(5x)x^{2} + 2 \cos{\left(5 x \right)} = x^{2} + 2 \cos{\left(5 x \right)}
- Yes
x2+2cos(5x)=x22cos(5x)x^{2} + 2 \cos{\left(5 x \right)} = - x^{2} - 2 \cos{\left(5 x \right)}
- No
so, the function
is
even