Monday, October 22, 2012
Practice:
Two consecutive extremes to f(x) = Asin(bx+c)+D
f(x) = Asin(bx+c)+D to a middle point and the nearest maximum and minimum points
Two consecutive extremes
a) (6, 3) and (10, 4)
b) (-2, 1) and (4, -1)
c) (3, 5) and (5, 3)
Find the midpoint and the nearest max and min for the following functions
f(x) = 3sin([pi/2]x + 4) - 5
g(x) = -sin(x + pi/4)
r(x) = -4sin(4x + 4) +4
answers in the comments.
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a) (6, 3) and (10, 4)
ReplyDeletemidpoint is (8, 3.5)
the first point is the min and the second the max, so A is positive and the amplitude is 3.5-3 = ½, so we begin with
f(x) = ½sin(bx+c) + 3½
The distance in x between the two extremes is |6-10|=4, so the period is 8. 8 = 2pi/b, so b=pi/4
To find c, solve 8pi/4 + c = 0, so c = -2pi.
f(x) = ½sin(pi/4*x-2pi) + 3½
b) (-2, 1) and (4, -1)
midpoint is (1, 0)
the first point is the max and the second the min, so A is negative and the amplitude is 0-1 = -1, so we begin with
f(x) = -sin(bx+c) + 0 = -sin(bx+c)
The distance in x between the two extremes is |-2-4|=6, so the period is 12. 12 = 2pi/b, so b=pi/6
To find c, solve pi/6 + c = 0, so c = -pi/6.
f(x) = -sin(pi/6x-pi/6)
c) (3, 5) and (5, 3)
midpoint is (4,4) max first and min second, so amplitude is negative, and 4-5 = -1
f(x) = -sin(bx+c) + 4
The distance between a max and min is 2 so the period = 4 = 2pi/b so b = pi/2. Since x = 4 at the midpoint, solve [pi/2]*4 + c = 0, so c = -2pi.
f(x) = -sin([pi/2]x - 2pi) + 4
Find the midpoint and the nearest max and min for the following functions
f(x) = 3sin([pi/2]x + 4) - 5
min = (___, -8)
middle = (___, -5)
max = (___, -2)
solve pi/2x+4 = 0, so x = -8/pi.
min = (___, -8)
middle = (-8/pi, -5)
max = (___, -2)
The period is 4, so the max and min are 1/4*4 away from the midpoint
min = (1-8/pi, -8)
middle = (-8/pi, -5)
max = (-1-8/pi, -2)
g(x) = -sin(x + pi/4)
min = (___, -1)
middle = (___, 0)
max = (___, 1)
solve x + pi/4 = 0 to get x = -pi/4
min = (___, -1)
middle = (-pi/4, 0)
max = (___, 1)
Period is 2pi, so max and min are pi/2 away from midpoint. Because A is negative the max is to the left and the min is to the right.
min = (pi/4, -1)
middle = (-pi/4, 0)
max = (-3pi/4, 1)
r(x) = -4sin(4x + 4) +4
min = (____, 0)
middle = (____, 4)
max = (____, 8)
solve 4x +4 = 0, x = -1.
min = (____, 0)
middle = (-1, 4)
max = (____, 8)
The period is 2pi/4 = pi/2, so the max and min happen at pi/8 away from the middle.
min = (-1+pi/8, 0)
middle = (-1, 4)
max = (-1-pi/8, 8)