Consider the function h(x)=-(2sin(1-x)+8). Giving your answer in interval notation, find the domain of h^(-1)(x).

Define h(x): The function h(x) is defined as h(x) = -(2sin(1-x) + 8). To find the domain of the inverse function h^(-1)(x), we first need to determine the range of the original function h(x), because the domain of h^(-1)(x) will be the range of h(x).Determine sine function range: The sine function oscillates between -1 and 1. Therefore, 2sin(1-x) will oscillate between -2 and 2. When we subtract this value from -8, the resulting values will oscillate between -8 - 2 = -10 and -8 + 2 = -6.Calculate h(x) range: Since h(x) = -(2sin(1-x) + 8), the range of h(x) is from -10 to -6, inclusive. This is because the sine function reaches its maximum and minimum values, and the negative sign in front of the function reflects the values.Find inverse function domain: The range of h(x) is the interval [-10, -6]. Therefore, the domain of the inverse function h^(-1)(x) is the same interval, [-10, -6].

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Consider the function h(x)=-(2sin(1-x)+8).
Giving your answer in interval notation, find the domain of h^(-1)(x).

Consider the function $h(x)=-(2 \sin (1-x)+8)$ . $\newline$ Giving your answer in interval notation, find the domain of $h^{-1}(x)$ .

View step-by-step help

Home

Math Problems

Algebra 1

Domain and range of quadratic functions: equations

Full solution

Q. Consider the function

h(x)=-(2 \sin (1-x)+8)

\newline

Giving your answer in interval notation, find the domain of

h^{-1}(x)

Define $h(x)$ : The function $h(x)$ is defined as $h(x) = -(2\sin(1-x) + 8)$ . To find the domain of the inverse function $h^{-1}(x)$ , we first need to determine the range of the original function $h(x)$ , because the domain of $h^{-1}(x)$ will be the range of $h(x)$ .
Determine sine function range: The sine function oscillates between $-1$ and $1$ . Therefore, $2\sin(1-x)$ will oscillate between $-2$ and $2$ . When we subtract this value from $-8$ , the resulting values will oscillate between $-8 - 2 = -10$ and $-8 + 2 = -6$ .
Calculate $h(x)$ range: Since $h(x) = -(2\sin(1-x) + 8)$ , the range of $h(x)$ is from $-10$ to $-6$ , inclusive. This is because the sine function reaches its maximum and minimum values, and the negative sign in front of the function reflects the values.
Find inverse function domain: The range of $h(x)$ is the interval $[-10, -6]$ . Therefore, the domain of the inverse function $h^{-1}(x)$ is the same interval, $[-10, -6]$ .