Question

The range of the function f(x) = 2sin(3x) +1 is equal to

• A. [-1, 1]
• B. \([\frac{-1}{3},\frac{1}{3}]\)
• C. [-2, 1]
• D. [-1, 2]
• E. [-1, 3]

Answer 1

Answer: (E)

The function given is \( f(x) = 2 \sin(3x) + 1 \). We know that the sine function \( \sin(x) \) has a range of \( [-1, 1] \).
Therefore, the expression \( 2 \sin(3x) \) will have a range of \( [-2, 2] \) because multiplying by 2 scales the range.
Now, adding 1 to this range shifts the entire range upward by 1 unit. Thus, the range of \( f(x) = 2 \sin(3x) + 1 \) is: \[ [-2 + 1, 2 + 1] = [-1, 3]. \]

The correct option is (E) : \([-1, 3]\)

Answer 2

We are given the function \(f(x) = 2\sin(3x) + 1\).

We know that the range of the sine function, \(\sin(x)\), is \([-1, 1]\). That is, \(-1 \le \sin(x) \le 1\) for all real numbers x.

Therefore, \(-1 \le \sin(3x) \le 1\).

Multiplying the inequality by 2, we get: \(2(-1) \le 2\sin(3x) \le 2(1)\), which simplifies to \(-2 \le 2\sin(3x) \le 2\).

Adding 1 to all parts of the inequality, we get: \(-2 + 1 \le 2\sin(3x) + 1 \le 2 + 1\), which simplifies to \(-1 \le 2\sin(3x) + 1 \le 3\).

Since \(f(x) = 2\sin(3x) + 1\), the range of \(f(x)\) is \([-1, 3]\).