S16 Given \( f(x)=\log _{\frac{1}{2}} x \). 7.2.1 Write down the equation of \( f^{-1} \), the inverse of \( f \), in the form \( y=\ldots \) 7.2.2 Draw a neat sketch graph of \( f^{-1} \). Clearly show all intercepts with the a: and any other point on the graph. Write down the equation of the function \( g \) if \( g \) is the mirror image of \( f \)-axis. 7.2.4 For which value(s) of \( x \) is: \( f(x) \cdot f^{-1}(x)<0 \) ?

To find \( f^{-1} \) given \( f(x)=\log _{\frac{1}{2}} x \), we start by setting \( y = f(x) \): 1. \( y = \log_{\frac{1}{2}} x \) can be rewritten as \( x = \left(\frac{1}{2}\right)^y \). 2. To solve for \( x \), we can also write this as \( x = 2^{-y} \). 3. Thus, the inverse function is \( f^{-1}(y) = 2^{-y} \) and written in the form \( y = f^{-1}(x) \), we have: \[ y = 2^{-x} \] For the mirror image of \( f \) across the x-axis, we simply negate the function: \( g(x) = -f(x) = -\log_{\frac{1}{2}} x \). Now, regarding the inequality \( f(x) \cdot f^{-1}(x)<0 \): 1. \( f(x) \) is negative when \( 0 < x < 1 \) (since the log base \(\frac{1}{2}\) is negative in this range). 2. \( f^{-1}(x) = 2^{-x} \) is positive for all \( x \), since the exponential function is always positive. Thus, the product \( f(x) \cdot f^{-1}(x) < 0 \) occurs when \( f(x) < 0 \) and \( f^{-1}(x) > 0\), specifically for \( 0 < x < 1 \).