Determine the absolute extreme values of the function on the interval .
- Thread starter JSmith
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There is more than one root of the equation sin(x) = -1/2.Determine the absolute extreme values of the functionon the interval
.
I know f'(x) = -2 sinx -1
0 = -2 sinx -1
1/2 = sinx X wrong sign
x = - pi / 6
Now what do I have to do? Do I have to do something with the radian circle, or is x=0.523 my critical value?
In addition, you have to consider the endpoint(s) of the interval. Evaluate f(x) at all the relevant points, and select the absolute min and max.
So I evaluate the function at 0, -pi/6, and 2pi?
0 and 2pi are the same. There is one more angle which has sinx = -1/2.
So -5pi/6?
D
Deleted member 4993
Guest
-5π/6 is not within the domain of 0 ≤ x ≤ 2π → cannot be one of the answers.
wait but neither is -pi/6... so do I only evaluate the end point?
D
Deleted member 4993
Guest
Look at the unit circle. In the domain 0 - 2π there two values of 'x' where sin(x) = -1/2
Sorry I didn't catch that I had given you a solution outside the range. You have to add 2pi to be correct. Remember all circular functions repeat every (2 n pi). Thus there are still two solutions of the equation
\(\displaystyle \sin{x} = - \frac{1}{2} \)
in the range [0, 2pi].
sin x=-1/2
x=(-1)^k*arcsin(-1/2) + k*pi
x= (-1)^(k+1)*arcsin(1/2) + k*pi
x= (-1)^(k+1)*(pi/6)+ k*pi
sine has negative values in the 3rd and 4th quadrants.
x1=pi + pi/6
x1=(6pi+pi)/6
x1=7pi/6
x2= 2pi- pi/6
x2=(12pi-pi)/6
x2 = 11pi/6
Correct?
x=(-1)^k*arcsin(-1/2) + k*pi
x= (-1)^(k+1)*arcsin(1/2) + k*pi
x= (-1)^(k+1)*(pi/6)+ k*pi
sine has negative values in the 3rd and 4th quadrants.
x1=pi + pi/6
x1=(6pi+pi)/6
x1=7pi/6
x2= 2pi- pi/6
x2=(12pi-pi)/6
x2 = 11pi/6
Correct?
D
Deleted member 4993
Guest
Correct - now check the values of 'y' at x1, x2, 0 and 2π
D
Deleted member 4993
Guest
Looks good. To check yourself, you may plot the function with a graphing calculator or visit wolframalpha.com