QUESTION 5 Given: \( f(x)=\left(\frac{1}{4}\right)^{x} \) 5.1 Determine the value of \( f(-2) \). 5.2 Write down the equation of \( f^{-1}(x) \) in the form \( y=\ldots \) (2) 5.3 Andrew has no idea how to draw the graph of \( f^{-1} \). Explain to Andrew how he may use the graph of \( f \) to draw the graph of \( f^{-1} \). 5.4 Hence or otherwise, sketch the graph of \( f^{-1} \). Clearly indicate ALL intercepts with the - axes. 5.5 Write down the domain of \( f^{-1} \). 5.6 For which values of \( x \) will \( f^{-1}(x) \geq-2 \) ? (2) (1) 5.7 Given: \( q=\log _{\frac{1}{4}} \frac{1}{2} \) 5.7.1 Determine the value of \( q \). (1) 5.7.2 Hence or otherwise, determine the coordinates of the point of intersection of \( f \) and \( f^{-1} \). [14]

To find the value of \( f(-2) \), substitute \(-2\) into the function: \[ f(-2) = \left(\frac{1}{4}\right)^{-2} = 4^2 = 16. \] Now, for the inverse function \( f^{-1}(x) \), we start with \( y = \left(\frac{1}{4}\right)^{x} \) and solve for \( x \): \[ x = \left(\frac{1}{4}\right)^{y} \Longrightarrow \log_{\frac{1}{4}}(x) = y \Longrightarrow f^{-1}(x) = \log_{\frac{1}{4}}(x). \] Andrew can utilize the graph of \( f \) to sketch \( f^{-1} \) by reflecting it across the line \( y = x \). Every point \((a, b)\) on the graph of \( f \) transforms to \((b, a)\) on the graph of \( f^{-1} \). This simple mirroring will show him the relationship between a function and its inverse visually. To sketch the graph of \( f^{-1} \), remember that the function is defined where \( x > 0 \) (since logarithms of non-positive numbers are undefined). The graph will pass through \((1, 0)\) since \( f(0) = 1 \) and \( f^{-1}(1) = 0\), and the intercept with the vertical axis at \( (0, -\infty) \). Horizontally, it approaches infinity as \( x \) nears zero. The domain of \( f^{-1} \) is all positive real numbers, i.e., \( (0, \infty) \). For the condition \( f^{-1}(x) \geq -2 \), set up the inequality: \[ \log_{\frac{1}{4}}(x) \geq -2 \Rightarrow x \leq \left(\frac{1}{4}\right)^{-2} = 16. \] Thus, \( f^{-1}(x) \geq -2 \) holds for \( 0 < x \leq 16 \). Now for the logarithmic value of \( q \): \[ q = \log_{\frac{1}{4}} \frac{1}{2} = \frac{\log \frac{1}{2}}{\log \frac{1}{4}} = \frac{-\frac{1}{2}}{-2} = \frac{1}{4}. \] The coordinates of the intersection of \( f \) and \( f^{-1} \) occur when \( f(x) = x \). Therefore, solving \( \left(\frac{1}{4}\right)^{x} = x \), you can find that the intersection happens roughly at \( x = 1 \), giving us the point \( (1, 1) \), which serves as a point of interest for both graphs.