the function f is given by f(x) = sin(2x), and the domain of f is -π/4 ≤ x ≤ π/4. which of the following…

the function f is given by f(x) = sin(2x), and the domain of f is -π/4 ≤ x ≤ π/4. which of the following defines f⁻¹(x)? a 1/2 arcsin x on the interval -1/2 ≤ x ≤ 1/2 b arcsin(1/2 x) on the interval -1/2 ≤ x ≤ 1/2 c 1/2 arcsin x on the interval -1 ≤ x ≤ 1 d arcsin(1/2 x) on the interval -1 ≤ x ≤ 1

the function f is given by f(x) = sin(2x), and the domain of f is -π/4 ≤ x ≤ π/4. which of the following defines f⁻¹(x)? a 1/2 arcsin x on the interval -1/2 ≤ x ≤ 1/2 b arcsin(1/2 x) on the interval -1/2 ≤ x ≤ 1/2 c 1/2 arcsin x on the interval -1 ≤ x ≤ 1 d arcsin(1/2 x) on the interval -1 ≤ x ≤ 1

Answer

Explanation:

Step1: Recall inverse - function definition

Let (y = f(x)=\sin(2x)). To find the inverse, we solve for (x) in terms of (y). First, rewrite the equation as (y=\sin(2x)). Then, by the definition of the inverse - sine function, (2x=\arcsin(y)) (since for (y = \sin(u)), (u=\arcsin(y)) when (-\frac{\pi}{2}\leq u\leq\frac{\pi}{2}). Here, when (x\in[-\frac{\pi}{4},\frac{\pi}{4}]), (2x\in[-\frac{\pi}{2},\frac{\pi}{2}])).

Step2: Solve for (x)

Dividing both sides of (2x = \arcsin(y)) by 2, we get (x=\frac{1}{2}\arcsin(y)). So, (f^{-1}(x)=\frac{1}{2}\arcsin(x)).

Step3: Find the domain of (f^{-1}(x))

The range of (y = f(x)=\sin(2x)) for (x\in[-\frac{\pi}{4},\frac{\pi}{4}]) is ([- 1,1]). The domain of the inverse function (f^{-1}(x)) is the range of the original function (f(x)). So the domain of (f^{-1}(x)) is (-1\leq x\leq1).

Answer:

C. (\frac{1}{2}\arcsin x) on the interval (-1\leq x\leq1)