f (x)=3sin4x-cos6x then find the difference of maximum and minimum

1.	f (x)=3sin4x-cos6x then find the difference of maximum and minimum
Answer» sin(x) and cos(x) both have range [-1,1], it should be obvious that the value of: sin²(x)∈[0,1] sin⁴(x)∈[0,1] 3sin⁴(x)∈[0,3] cos²(x)∈[0,1] cos⁶(x)∈[0,1] The max^m value of 3sin⁴(x) =3 (when x = (2n+1/2)π ) The mini^m value of 3sin⁴(x) = 0 (when x = nπ ) The max^m value of cos⁶(x) =1 (when x = nπ ) The mini^m value of cos⁶(x) = 0 (when x = (2n+1/2)π ) So, at x=nπ, 3sin⁴(x) has its minimum value and cos⁶(x)has its maximum value, so for f(x) to be minimum value (= 0–1 = -1). When x= (2n + 1/2)π, 3sin⁴(x) has its maximum value and cos⁶(x)has its minimum value, so for f(x) to be maximum value (= 3–0 = 3). Hence, the difference between the maximum and minimum values of f(x) = 4

Answer»

sin(x) and cos(x) both have range [-1,1], it should be obvious that the value of:

sin²(x)∈[0,1]

sin⁴(x)∈[0,1]

3sin⁴(x)∈[0,3]

cos²(x)∈[0,1]

cos⁶(x)∈[0,1]

The max^m value of 3sin⁴(x) =3 (when x = (2n+1/2)π )
The mini^m value of 3sin⁴(x) = 0 (when x = nπ )

The max^m value of cos⁶(x) =1 (when x = nπ )
The mini^m value of cos⁶(x) = 0 (when x = (2n+1/2)π )

So, at x=nπ, 3sin⁴(x) has its minimum value and cos⁶(x)has its maximum value,
so for f(x) to be minimum value (= 0–1 = -1).

When x= (2n + 1/2)π, 3sin⁴(x) has its maximum value and cos⁶(x)has its minimum value,
so for f(x) to be maximum value (= 3–0 = 3).

Hence, the difference between the maximum and minimum values of f(x) = 4

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