Mathematics Progress check Sub What is the inverse of the following function? \[ f(x)=(x-9)^{2} \] A. \( m(x)= \pm \sqrt{x-9} \), for \( x \geq 0 \) B. \( n(x)=\sqrt{x \pm 9} \), for \( x \geq 0 \) C. \( p(x)=9 \pm \sqrt{x} \), for \( x \geq 0 \) D. \( q(x)=\sqrt{x} \pm 9 \), for \( x \geq 0 \) Reset Next

To find the inverse of the function \( f(x)=(x-9)^{2} \), we first need to solve for \( x \) in terms of \( y \). Setting \( y = (x-9)^{2} \), we take the square root of both sides, yielding \( \sqrt{y} = x - 9 \). Thus, \( x = \sqrt{y} + 9 \). Since the original function is a squared term, it is not one-to-one across its entire domain, but we can restrict it to \( x \geq 9 \) to ensure the inverse is also a function. Therefore, the inverse is \( f^{-1}(x) = \sqrt{x} + 9 \), which means we should look for the correct corresponding option. The correct answer is: none of the options exactly match the derived inverse, but the reasoning can help in identifying how inverses work for quadratic functions! Another fun fact is that when dealing with functions like this, understanding the concept of domain and range is crucial. The range of \( f(x) \) starts at 0 (since any real number squared is 0 or positive) and extends to positive infinity. Consequently, the domain of the inverse must match, allowing only non-negative outputs for the square root, which is why we restrict our domain and range accordingly!