Find the range of f(x) = 1/(1-2cosx)
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Given, f(x) = 1/(1 - 2cosx)
For defining the function , 1 - 2cosx ≠ 0
we assume 1 - 2cosx = 0.
Then, 1/2 = cosx ⇒x = π/3 and 5π/3
It means, in domain x ≠ π/3 and 5π/3
We consider interval ( 0, 2π)
Then, domain ∈(0,π/3) ∪ (π/3 , 5π/3) ∪ (5π/3 , 2π)
We know range is written as [least value, greatest value ]
so, let's take value of f(x) in each interval .
When x = 0 we have f(0) = 1/(1 -2cos(0)) = -1 it is maximum range of f(x)
So, range always less than -1
y ≤ -1
Now, let's check f(π) = 1/(1 - 2cosπ) = 1/3
Because π Is in the interval (π/3, 5π/3), and it is minimum range of f(x)
so, range of f(x) always greater than 1/3
Finally, range ∈(-∞ , -1] U [1/3, ∞)
For defining the function , 1 - 2cosx ≠ 0
we assume 1 - 2cosx = 0.
Then, 1/2 = cosx ⇒x = π/3 and 5π/3
It means, in domain x ≠ π/3 and 5π/3
We consider interval ( 0, 2π)
Then, domain ∈(0,π/3) ∪ (π/3 , 5π/3) ∪ (5π/3 , 2π)
We know range is written as [least value, greatest value ]
so, let's take value of f(x) in each interval .
When x = 0 we have f(0) = 1/(1 -2cos(0)) = -1 it is maximum range of f(x)
So, range always less than -1
y ≤ -1
Now, let's check f(π) = 1/(1 - 2cosπ) = 1/3
Because π Is in the interval (π/3, 5π/3), and it is minimum range of f(x)
so, range of f(x) always greater than 1/3
Finally, range ∈(-∞ , -1] U [1/3, ∞)
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Answer:
range of cos x is [-1,1]
from this you can derive range of f of x
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