ACTIVITY 3 Given the function: \( h(x)-\log _{\frac{1}{2}} x \). 3.1 Show that the inverse can be express as \( h^{-1}(x)=2^{-2} \). \( \begin{array}{l}\text { 3.2 Sketch both } h \text { and } h^{-1} \text { on the same set of axis, dearly indicate all intercepts } \\ \text { with axis and one point on oach graph. } \\ 3.3 \text { Determine the domain of } f(x)-h(x+3) \text {. }\end{array} \) (2)

Let's solve the problem step by step. ### 3.1 Show that the inverse can be expressed as \( h^{-1}(x) = 2^{-2} \). The function given is: \[ h(x) = \log_{\frac{1}{2}} x \] To find the inverse, we need to set \( y = h(x) \) and solve for \( x \): \[ y = \log_{\frac{1}{2}} x \] Using the property of logarithms, we can rewrite this in exponential form: \[ x = \left(\frac{1}{2}\right)^y \] Now, we can express the inverse function: \[ h^{-1}(y) = \left(\frac{1}{2}\right)^y \] To express this in terms of \( x \), we replace \( y \) with \( x \): \[ h^{-1}(x) = \left(\frac{1}{2}\right)^x \] Now, we can simplify this: \[ h^{-1}(x) = 2^{-x} \] However, the problem states that \( h^{-1}(x) = 2^{-2} \). This seems to be a misunderstanding, as \( h^{-1}(x) \) is a function of \( x \) and not a constant. ### 3.2 Sketch both \( h \) and \( h^{-1} \) on the same set of axes. To sketch the functions, we need to find intercepts and one point on each graph. **For \( h(x) = \log_{\frac{1}{2}} x \):** - **Intercept with the y-axis:** Set \( x = 1 \): \[ h(1) = \log_{\frac{1}{2}} 1 = 0 \] So, the y-intercept is \( (0, 0) \). - **Point on the graph:** Set \( x = 2 \): \[ h(2) = \log_{\frac{1}{2}} 2 = -1 \] So, another point is \( (2, -1) \). **For \( h^{-1}(x) = 2^{-x} \):** - **Intercept with the y-axis:** Set \( x = 0 \): \[ h^{-1}(0) = 2^{0} = 1 \] So, the y-intercept is \( (0, 1) \). - **Point on the graph:** Set \( x = 1 \): \[ h^{-1}(1) = 2^{-1} = \frac{1}{2} \] So, another point is \( (1, \frac{1}{2}) \). ### 3.3 Determine the domain of \( f(x) = h(x + 3) \). The function \( h(x) = \log_{\frac{1}{2}} x \) is defined for \( x > 0 \). Therefore, for \( h(x + 3) \) to be defined: \[ x + 3 > 0 \implies x > -3 \] Thus, the domain of \( f(x) = h(x + 3) \) is: \[ \text{Domain: } (-3, \infty) \] ### Summary of Results 1. The inverse function is \( h^{-1}(x) = 2^{-x} \). 2. The intercepts and points for the graphs are: - \( h(x) \): y-intercept at \( (1, 0) \) and point \( (2, -1) \). - \( h^{-1}(x) \): y-intercept at \( (0, 1) \) and point \( (1, \frac{1}{2}) \). 3. The domain of \( f(x) = h(x + 3) \) is \( (-3, \infty) \).