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Plotting a function graph/ y=0,5sinx+2

Graphing y = y=0,5sinx+2

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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       sin(x)    
f(x) = ------ + 2
         2
f(x)=sin⁡(x)2+2f{\left(x \right)} = \frac{\sin{\left(x \right)}}{2} + 2 
f = sin(x)/2 + 2
The graph of the function
02468-8-6-4-2-101013
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:
sin⁡(x)2+2=0\frac{\sin{\left(x \right)}}{2} + 2 = 0
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to sin(x)/2 + 2.
sin⁡(0)2+2\frac{\sin{\left(0 \right)}}{2} + 2
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
cos⁡(x)2=0\frac{\cos{\left(x \right)}}{2} = 0
Solve this equation
The roots of this equation
x1=π2x_{1} = \frac{\pi}{2}
x2=3π2x_{2} = \frac{3 \pi}{2}
The values of the extrema at the points:
 pi      
(--, 5/2)
 2       

 3*pi      
(----, 3/2)
  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=3π2x_{1} = \frac{3 \pi}{2}
Maxima of the function at points:
x1=π2x_{1} = \frac{\pi}{2}
Decreasing at intervals
(−∞,π2]∪[3π2,∞)\left(-\infty, \frac{\pi}{2}\right] \cup \left[\frac{3 \pi}{2}, \infty\right)
Increasing at intervals
[π2,3π2]\left[\frac{\pi}{2}, \frac{3 \pi}{2}\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
−sin⁡(x)2=0- \frac{\sin{\left(x \right)}}{2} = 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(-\infty, 0\right] \cup \left[\pi, \infty\right)
Convex at the intervals
[0,π]\left[0, \pi\right]
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(sin⁡(x)2+2)=⟨32,52⟩\lim_{x \to -\infty}\left(\frac{\sin{\left(x \right)}}{2} + 2\right) = \left\langle \frac{3}{2}, \frac{5}{2}\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=⟨32,52⟩y = \left\langle \frac{3}{2}, \frac{5}{2}\right\rangle
lim⁡x→∞(sin⁡(x)2+2)=⟨32,52⟩\lim_{x \to \infty}\left(\frac{\sin{\left(x \right)}}{2} + 2\right) = \left\langle \frac{3}{2}, \frac{5}{2}\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=⟨32,52⟩y = \left\langle \frac{3}{2}, \frac{5}{2}\right\rangle
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of sin(x)/2 + 2, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(sin⁡(x)2+2x)=0\lim_{x \to -\infty}\left(\frac{\frac{\sin{\left(x \right)}}{2} + 2}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(sin⁡(x)2+2x)=0\lim_{x \to \infty}\left(\frac{\frac{\sin{\left(x \right)}}{2} + 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:
sin⁡(x)2+2=2−sin⁡(x)2\frac{\sin{\left(x \right)}}{2} + 2 = 2 - \frac{\sin{\left(x \right)}}{2}
- No
sin⁡(x)2+2=sin⁡(x)2−2\frac{\sin{\left(x \right)}}{2} + 2 = \frac{\sin{\left(x \right)}}{2} - 2
- No
so, the function
not is
neither even, nor odd
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