Mister Exam

Other calculators

Plotting a function graph/ x-2*sin(x+1/2)

Graphing y = x-2*sin(x+1/2)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
f(x) = x - 2*sin(x + 1/2)
f(x)=x−2sin⁡(x+12)f{\left(x \right)} = x - 2 \sin{\left(x + \frac{1}{2} \right)} 
f = x - 2*sin(x + 1/2)
The graph of the function
02468-8-6-4-2-1010-2525
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:
x−2sin⁡(x+12)=0x - 2 \sin{\left(x + \frac{1}{2} \right)} = 0
Solve this equation
The points of intersection with the axis X:

Numerical solution
x1=−1.05659293573137x_{1} = -1.05659293573137
x2=1.66130979346545x_{2} = 1.66130979346545
x3=−1.99412427587496x_{3} = -1.99412427587496
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to x - 2*sin(x + 1/2).
−2sin⁡(12)- 2 \sin{\left(\frac{1}{2} \right)}
The result:
f(0)=−2sin⁡(12)f{\left(0 \right)} = - 2 \sin{\left(\frac{1}{2} \right)}
The point:
(0, -2*sin(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
1−2cos⁡(x+12)=01 - 2 \cos{\left(x + \frac{1}{2} \right)} = 0
Solve this equation
The roots of this equation
x1=−12+π3x_{1} = - \frac{1}{2} + \frac{\pi}{3}
x2=−12+5π3x_{2} = - \frac{1}{2} + \frac{5 \pi}{3}
The values of the extrema at the points:
   1   pi    1     ___   pi 
(- - + --, - - - \/ 3  + --)
   2   3     2           3  

   1   5*pi    1     ___   5*pi 
(- - + ----, - - + \/ 3  + ----)
   2    3      2            3


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=−12+π3x_{1} = - \frac{1}{2} + \frac{\pi}{3}
Maxima of the function at points:
x1=−12+5π3x_{1} = - \frac{1}{2} + \frac{5 \pi}{3}
Decreasing at intervals
[−12+π3,−12+5π3]\left[- \frac{1}{2} + \frac{\pi}{3}, - \frac{1}{2} + \frac{5 \pi}{3}\right]
Increasing at intervals
(−∞,−12+π3]∪[−12+5π3,∞)\left(-\infty, - \frac{1}{2} + \frac{\pi}{3}\right] \cup \left[- \frac{1}{2} + \frac{5 \pi}{3}, \infty\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
2sin⁡(x+12)=02 \sin{\left(x + \frac{1}{2} \right)} = 0
Solve this equation
The roots of this equation
x1=−12x_{1} = - \frac{1}{2}
x2=−12+πx_{2} = - \frac{1}{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
[−12,−12+π]\left[- \frac{1}{2}, - \frac{1}{2} + \pi\right]
Convex at the intervals
(−∞,−12]∪[−12+π,∞)\left(-\infty, - \frac{1}{2}\right] \cup \left[- \frac{1}{2} + \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→−∞(x−2sin⁡(x+12))=−∞\lim_{x \to -\infty}\left(x - 2 \sin{\left(x + \frac{1}{2} \right)}\right) = -\infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞(x−2sin⁡(x+12))=∞\lim_{x \to \infty}\left(x - 2 \sin{\left(x + \frac{1}{2} \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 x - 2*sin(x + 1/2), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(x−2sin⁡(x+12)x)=1\lim_{x \to -\infty}\left(\frac{x - 2 \sin{\left(x + \frac{1}{2} \right)}}{x}\right) = 1
Let's take the limit
so,
inclined asymptote equation on the left:
y=xy = x
lim⁡x→∞(x−2sin⁡(x+12)x)=1\lim_{x \to \infty}\left(\frac{x - 2 \sin{\left(x + \frac{1}{2} \right)}}{x}\right) = 1
Let's take the limit
so,
inclined asymptote equation on the right:
y=xy = 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:
x−2sin⁡(x+12)=−x+2sin⁡(x−12)x - 2 \sin{\left(x + \frac{1}{2} \right)} = - x + 2 \sin{\left(x - \frac{1}{2} \right)}
- No
x−2sin⁡(x+12)=x−2sin⁡(x−12)x - 2 \sin{\left(x + \frac{1}{2} \right)} = x - 2 \sin{\left(x - \frac{1}{2} \right)}
- No
so, the function
not is
neither even, nor odd
    © Mister Exam – Calculator