Mister Exam

Other calculators

  • How to use it?

  • Graphing y =:
  • x^3/(2(x+1)^2)
  • x^2*(x-2)^2
  • x/(1+x)^2
  • (x^3+2x^2)/(x-1)^2

  • Identical expressions

  • sin(x)^ two - two *sin(x)+ one
  • sinus of (x) squared minus 2 multiply by sinus of (x) plus 1
  • sinus of (x) to the power of two minus two multiply by sinus of (x) plus one
  • sin(x)2-2*sin(x)+1
  • sinx2-2*sinx+1
  • sin(x)²-2*sin(x)+1
  • sin(x) to the power of 2-2*sin(x)+1
  • sin(x)^2-2sin(x)+1
  • sin(x)2-2sin(x)+1
  • sinx2-2sinx+1
  • sinx^2-2sinx+1

  • Similar expressions

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

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

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
          2                  
f(x) = sin (x) - 2*sin(x) + 1
f(x)=(sin⁡2(x)−2sin⁡(x))+1f{\left(x \right)} = \left(\sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)}\right) + 1 
f = sin(x)^2 - 2*sin(x) + 1
The graph of the function
02468-8-6-4-2-101005
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⁡2(x)−2sin⁡(x))+1=0\left(\sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)}\right) + 1 = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=π2x_{1} = \frac{\pi}{2}
Numerical solution
x1=70.6846901522759x_{1} = 70.6846901522759
x2=−29.8448001384705x_{2} = -29.8448001384705
x3=−73.8271871062847x_{3} = -73.8271871062847
x4=95.8191678819566x_{4} = 95.8191678819566
x5=−67.5448127096766x_{5} = -67.5448127096766
x6=−23.5624688735391x_{6} = -23.5624688735391
x7=58.1200311840253x_{7} = 58.1200311840253
x8=20.4198754083154x_{8} = 20.4198754083154
x9=1.57204027951788x_{9} = 1.57204027951788
x10=−17.2799879444777x_{10} = -17.2799879444777
x11=7.85446803582028x_{11} = 7.85446803582028
x12=−48.695589715671x_{12} = -48.695589715671
x13=45.5543801721935x_{13} = 45.5543801721935
x14=−61.2623265905076x_{14} = -61.2623265905076
x15=−86.3926407020155x_{15} = -86.3926407020155
x16=26.7023509688056x_{16} = 26.7023509688056
x17=−42.4103000063203x_{17} = -42.4103000063203
x18=64.4022201556448x_{18} = 64.4022201556448
x19=−80.1102366603878x_{19} = -80.1102366603878
x20=−4.71386903378524x_{20} = -4.71386903378524
x21=−92.6773772080111x_{21} = -92.6773772080111
x22=89.5367196861079x_{22} = 89.5367196861079
x23=51.8368185218287x_{23} = 51.8368185218287
x24=−36.1278836651147x_{24} = -36.1278836651147
x25=14.1376294337918x_{25} = 14.1376294337918
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(x) + 1.
(sin⁡2(0)−2sin⁡(0))+1\left(\sin^{2}{\left(0 \right)} - 2 \sin{\left(0 \right)}\right) + 1
The result:
f(0)=1f{\left(0 \right)} = 1
The point:
(0, 1)
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)cos⁡(x)−2cos⁡(x)=02 \sin{\left(x \right)} \cos{\left(x \right)} - 2 \cos{\left(x \right)} = 0
Solve this equation
The roots of this equation
x1=−π2x_{1} = - \frac{\pi}{2}
x2=π2x_{2} = \frac{\pi}{2}
The values of the extrema at the points:
 -pi     
(----, 4)
  2      

 pi    
(--, 0)
 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:
The function has no minima
Maxima of the function at points:
x2=−π2x_{2} = - \frac{\pi}{2}
Decreasing at intervals
(−∞,−π2]\left(-\infty, - \frac{\pi}{2}\right]
Increasing at intervals
[−π2,∞)\left[- \frac{\pi}{2}, \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
2(−sin⁡2(x)+sin⁡(x)+cos⁡2(x))=02 \left(- \sin^{2}{\left(x \right)} + \sin{\left(x \right)} + \cos^{2}{\left(x \right)}\right) = 0
Solve this equation
The roots of this equation
x1=−5π6x_{1} = - \frac{5 \pi}{6}
x2=−π6x_{2} = - \frac{\pi}{6}
x3=π2x_{3} = \frac{\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
(−∞,−5π6]∪[−π6,∞)\left(-\infty, - \frac{5 \pi}{6}\right] \cup \left[- \frac{\pi}{6}, \infty\right)
Convex at the intervals
[−5π6,−π6]\left[- \frac{5 \pi}{6}, - \frac{\pi}{6}\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⁡2(x)−2sin⁡(x))+1)=⟨−1,4⟩\lim_{x \to -\infty}\left(\left(\sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)}\right) + 1\right) = \left\langle -1, 4\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=⟨−1,4⟩y = \left\langle -1, 4\right\rangle
lim⁡x→∞((sin⁡2(x)−2sin⁡(x))+1)=⟨−1,4⟩\lim_{x \to \infty}\left(\left(\sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)}\right) + 1\right) = \left\langle -1, 4\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=⟨−1,4⟩y = \left\langle -1, 4\right\rangle
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of sin(x)^2 - 2*sin(x) + 1, divided by x at x->+oo and x ->-oo
lim⁡x→−∞((sin⁡2(x)−2sin⁡(x))+1x)=0\lim_{x \to -\infty}\left(\frac{\left(\sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)}\right) + 1}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞((sin⁡2(x)−2sin⁡(x))+1x)=0\lim_{x \to \infty}\left(\frac{\left(\sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)}\right) + 1}{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⁡2(x)−2sin⁡(x))+1=sin⁡2(x)+2sin⁡(x)+1\left(\sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)}\right) + 1 = \sin^{2}{\left(x \right)} + 2 \sin{\left(x \right)} + 1
- No
(sin⁡2(x)−2sin⁡(x))+1=−sin⁡2(x)−2sin⁡(x)−1\left(\sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)}\right) + 1 = - \sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)} - 1
- No
so, the function
not is
neither even, nor odd
    © Mister Exam – Calculator