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Assuming that x is in the appropriate domain, find a formula for h^(-1)(x)=1-(sin^(-1)(((-8-x))/(2)))

Assuming that x x is in the appropriate domain, find a formula for\newlineh−1(x)=1−(sin⁡−1((−8−x)2)) h^{-1}(x)=1-\left(\sin ^{-1}\left(\frac{(-8-x)}{2}\right)\right)

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Full solution

Q. Assuming that x x is in the appropriate domain, find a formula for\newlineh−1(x)=1−(sin⁡−1((−8−x)2)) h^{-1}(x)=1-\left(\sin ^{-1}\left(\frac{(-8-x)}{2}\right)\right)
  1. Start with given equation: To find the formula for the inverse function h−1(x)h^{-1}(x), we need to express xx in terms of hh. We start with the given equation for h−1(x)h^{-1}(x):\newlineh−1(x)=1−sin⁡−1(−8−x2)h^{-1}(x) = 1 - \sin^{-1}\left(\frac{-8 - x}{2}\right)\newlineLet's denote h−1(x)h^{-1}(x) as yy for simplicity:\newliney=1−sin⁡−1(−8−x2)y = 1 - \sin^{-1}\left(\frac{-8 - x}{2}\right)
  2. Isolate sin⁡−1\sin^{-1} term: Now we need to solve for xx. First, we isolate the sin⁡−1\sin^{-1} term by moving everything else to the other side of the equation:\newlinesin⁡−1(−8−x2)=1−y\sin^{-1}\left(\frac{-8 - x}{2}\right) = 1 - y
  3. Apply sine function: Next, we apply the sine function to both sides to remove the inverse sine, keeping in mind that sin⁡(sin⁡−1(u))=u\sin(\sin^{-1}(u)) = u for all uu in the domain of sin⁡−1\sin^{-1}:sin⁡(sin⁡−1(−8−x2))=sin⁡(1−y)\sin(\sin^{-1}(\frac{-8 - x}{2})) = \sin(1 - y)−8−x2=sin⁡(1−y)\frac{-8 - x}{2} = \sin(1 - y)
  4. Multiply by 22: Now we multiply both sides by 22 to get rid of the denominator:\newline−8−x=2sin⁡(1−y)-8 - x = 2\sin(1 - y)
  5. Isolate xx: Next, we isolate xx by moving the −8-8 to the other side:\newline−x=2sin⁡(1−y)+8-x = 2\sin(1 - y) + 8
  6. Multiply by −1-1: Finally, we multiply both sides by −1-1 to solve for xx:x=−2sin⁡(1−y)−8x = -2\sin(1 - y) - 8
  7. Replace yy with h−1(x)h^{-1}(x): Now we replace yy with h−1(x)h^{-1}(x) to get the formula for the inverse function in terms of xx:
    x=−2sin⁡(1−h−1(x))−8x = -2\sin(1 - h^{-1}(x)) - 8
    However, this is not quite correct because we need to express h−1(x)h^{-1}(x) in terms of xx, not the other way around. We need to go back and correct this mistake.

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