4* 1 paint What is the inverse of \( h(x)=7 x^{2}+8 \) with the domain restriction \( x \geq 0 \) ? A \( h^{-1}(x)=\sqrt{\frac{x-1}{9}} \) B. \( h^{-1}(x)=\frac{\sqrt{x-1}}{h} \) c \( h^{-1}(x)=\frac{\sqrt{x+1}}{y} \) D. \( x^{-1}(x)=\sqrt{x-7}+8 \) A B C D)

To find the inverse of the function \( h(x) = 7x^2 + 8 \) with a domain restriction of \( x \geq 0 \), we start by setting \( y = h(x) \). This gives us the equation \( y = 7x^2 + 8 \). To find the inverse, we solve for \( x \): \[ y - 8 = 7x^2 \] \[ x^2 = \frac{y - 8}{7} \] Since \( x \) must be non-negative due to our domain restriction, we take the positive square root: \[ x = \sqrt{\frac{y - 8}{7}} \] This gives us the inverse function: \[ h^{-1}(y) = \sqrt{\frac{y - 8}{7}} \] Substituting back to use \( x \) instead of \( y \), we have: \[ h^{-1}(x) = \sqrt{\frac{x - 8}{7}} \] Now, looking at your answer choices, it seems none directly match this expression. However, option A seems like it might have a typo, as it lists '1' in place of '8'. Make sure to check that answer for accuracy. When solving equations involving inverse functions, a common mistake is forgetting to switch the variables correctly after finding the expression; always ensure the notation aligns! Another is failing to account for domain restrictions; remember that if \( x \) was initially restricted, \( h^{-1}(x) \) must also respect changes in values allowed across the domains.