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x (cos^1) x question (1 Viewer)
- Thread starter akuchan
- Start date
b0b444
Member
- Joined
- Nov 4, 2004
- Messages
- 50
- Gender
- Male
- HSC
- 2005
hmmm, this is an interesting one. i can't give you a deadset method that you can learn whenever anything like this comes up. all i can give you is some logical thought.
the first thing to note is the domain of the function.
x.cos(^-1)x
it has to be -1<x<1. inverse cos does not exist outside these boundaries. prove this to yourself by trying to sub in x=2. what happens?
now sub in a few values of x in this domain.
x=1: 1 x inversecos(1) = 0
x=1/2: 1/2 x inversecos(1/2) = 1/2 x pi/3 = pi/6
x=0: 0 x inversecos(0) = 0
the interesting things start happening when you start to sub in negative numbers. the x out the front means that the y values of the function will definitely be negative (since inverse cos is never negative). sub in more numbers and eventually you get.
x=-1: -1 x inversecos(-1) = -1 x pi = -pi
at this point its reasonable to assume that since inversecos(x) and x are both continuous functions with no stationary points then x.inversecos(x) is the same so the maximum and minimum you've gotten so far are the maximum and minimum of the range.
therefore range is -pi<= y <= pi/6.
not what you said.
...i bet thats as clear as mud
the first thing to note is the domain of the function.
x.cos(^-1)x
it has to be -1<x<1. inverse cos does not exist outside these boundaries. prove this to yourself by trying to sub in x=2. what happens?
now sub in a few values of x in this domain.
x=1: 1 x inversecos(1) = 0
x=1/2: 1/2 x inversecos(1/2) = 1/2 x pi/3 = pi/6
x=0: 0 x inversecos(0) = 0
the interesting things start happening when you start to sub in negative numbers. the x out the front means that the y values of the function will definitely be negative (since inverse cos is never negative). sub in more numbers and eventually you get.
x=-1: -1 x inversecos(-1) = -1 x pi = -pi
at this point its reasonable to assume that since inversecos(x) and x are both continuous functions with no stationary points then x.inversecos(x) is the same so the maximum and minimum you've gotten so far are the maximum and minimum of the range.
therefore range is -pi<= y <= pi/6.
not what you said.
...i bet thats as clear as mud