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Plotting a function graph/ 2x-sinx

Graphing y = 2x-sinx

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The graph:

from to

Intersection points:

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Piecewise:

The solution

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f(x) = 2*x - sin(x)
f(x)=2x−sin⁡(x)f{\left(x \right)} = 2 x - \sin{\left(x \right)} 
f = 2*x - sin(x)
The graph of the function
-2.0-1.8-1.6-1.4-1.2-1.0-0.8-0.6-0.4-0.20.05-5
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:
2x−sin⁡(x)=02 x - \sin{\left(x \right)} = 0
Solve this equation
The points of intersection with the axis X:

Numerical solution
x1=0x_{1} = 0
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 2*x - sin(x).
0⋅2−sin⁡(0)0 \cdot 2 - \sin{\left(0 \right)}
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
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
2−cos⁡(x)=02 - \cos{\left(x \right)} = 0
Solve this equation
Solutions are not found,
function may have no extrema
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
sin⁡(x)=0\sin{\left(x \right)} = 0
Solve this equation
The roots of this equation
x1=0x_{1} = 0
x2=πx_{2} = \pi

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
[0,π]\left[0, \pi\right]
Convex at the intervals
(−∞,0]∪[π,∞)\left(-\infty, 0\right] \cup \left[\pi, \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→−∞(2x−sin⁡(x))=−∞\lim_{x \to -\infty}\left(2 x - \sin{\left(x \right)}\right) = -\infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞(2x−sin⁡(x))=∞\lim_{x \to \infty}\left(2 x - \sin{\left(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*x - sin(x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(2x−sin⁡(x)x)=2\lim_{x \to -\infty}\left(\frac{2 x - \sin{\left(x \right)}}{x}\right) = 2
Let's take the limit
so,
inclined asymptote equation on the left:
y=2xy = 2 x
lim⁡x→∞(2x−sin⁡(x)x)=2\lim_{x \to \infty}\left(\frac{2 x - \sin{\left(x \right)}}{x}\right) = 2
Let's take the limit
so,
inclined asymptote equation on the right:
y=2xy = 2 x
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:
2x−sin⁡(x)=−2x+sin⁡(x)2 x - \sin{\left(x \right)} = - 2 x + \sin{\left(x \right)}
- No
2x−sin⁡(x)=2x−sin⁡(x)2 x - \sin{\left(x \right)} = 2 x - \sin{\left(x \right)}
- No
so, the function
not is
neither even, nor odd
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