Example 1 :
Find the domain and range of the following functions
f(x) = 1/(2 - sin 3x)
Solution :
In order to find the domain, let us equate the denominator to 0.
2 - sin 3x = 0
sin 3x = 2
This is impossible, because the minimum value of sin is -1 and maximum value is 1. So we will not the above situation at any more.
So, the domain is all real values.
Range for sin function is between -1 and 1.
-1 ≤ sin 3x ≤ 1
Multiply by negative sign, through out
1 ≤ -sin 3x ≤ -1
-1 ≤ -sin 3x ≤ 1
Adding 2 through out, we get
-1 + 2 ≤ -sin 3x + 2 ≤ 1 + 2
1 ≤ 2 - sin 3x ≤ 3
3 ≤ (2 - sin 3x) ≤ 1
Taking reciprocals, we get
(1/3) ≤ 1/(2 - sin 3x) ≤ (1/1)
(1/3) ≤ 1/(2 - sin 3x) ≤ 1
So, the range is [1/3, 1]
Example 2 :
Find the range of the following function.
f(x) = 1/(1 - 2 cos x)
Solution :
Generally range for the cos function lies between -1 and 1
-1 ≤ cos x ≤ 1
Multiply by -2 throughout, we get
2 ≤ -2 cos x ≤ -2
-2 ≤ -2 cos x ≤ 2
Add 1 though out the line
(-2 + 1) ≤ (-2 cos x + 1) ≤ (2 + 1)
-1 ≤ 1 - 2 cos x ≤ 3
Taking reciprocal, we get
-1/1 ≤ 1/(1 - 2 cos x) ≤ (1/3)
-1 ≤ 1/(1 - 2 cos x) ≤ (1/3)
So, the range is [-1, 1/3]
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