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Consider the function h(x)=(2sin(1x)+8)h(x) = -(2\sin(1-x) + 8). \newlineGiving your answer in interval notation, find the domain of h1(x)h^{-1}(x).

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Q. Consider the function h(x)=(2sin(1x)+8)h(x) = -(2\sin(1-x) + 8). \newlineGiving your answer in interval notation, find the domain of h1(x)h^{-1}(x).
  1. Define h(x)h(x): The function h(x)h(x) is defined as h(x)=(2sin(1x)+8)h(x) = -(2\sin(1-x) + 8). To find the domain of the inverse function h1(x)h^{-1}(x), we first need to determine the range of the original function h(x)h(x), because the domain of h1(x)h^{-1}(x) will be the range of h(x)h(x).
  2. Determine sine function range: The sine function oscillates between 1-1 and 11. Therefore, 2sin(1x)2\sin(1-x) will oscillate between 2-2 and 22. When we subtract this from 8-8, the resulting values will oscillate between 82=10-8 - 2 = -10 and 8+2=6-8 + 2 = -6.
  3. Calculate h(x)h(x) range: Since h(x)=(2sin(1x)+8)h(x) = -(2\sin(1-x) + 8), the range of h(x)h(x) is from 10-10 to 6-6, 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.
  4. Find inverse function domain: The range of h(x)h(x) is the interval [10,6][-10, -6]. Therefore, the domain of the inverse function h1(x)h^{-1}(x) is the same interval, [10,6][-10, -6].

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