Werse of \( y=a x^{-2} \) Given the function \( f(x)=x^{2} \) 5.1 Determine \( f^{-1}(x) \). 5.2 Draw the graph of \( f^{-1}(x) \). 5.3 Explain why \( f^{-1}(x) \) will not be a function? 5.4 Explain how you will restrict the domain of \( f(x) \) to ensure that \( f^{-1}(x) \) wil be a function.

Let's solve the problem step by step. ### 5.1 Determine \( f^{-1}(x) \) The function given is \( f(x) = x^2 \). To find the inverse function \( f^{-1}(x) \), we follow these steps: 1. **Replace \( f(x) \) with \( y \)**: \[ y = x^2 \] 2. **Swap \( x \) and \( y \)**: \[ x = y^2 \] 3. **Solve for \( y \)**: \[ y = \sqrt{x} \quad \text{or} \quad y = -\sqrt{x} \] Thus, the inverse function can be expressed as: \[ f^{-1}(x) = \sqrt{x} \quad \text{and} \quad f^{-1}(x) = -\sqrt{x} \] However, since we need a single-valued function, we will choose one branch. Typically, we take the positive branch: \[ f^{-1}(x) = \sqrt{x} \] ### 5.2 Draw the graph of \( f^{-1}(x) \) To graph \( f^{-1}(x) = \sqrt{x} \), we can plot points for various values of \( x \): - For \( x = 0 \), \( f^{-1}(0) = 0 \) - For \( x = 1 \), \( f^{-1}(1) = 1 \) - For \( x = 4 \), \( f^{-1}(4) = 2 \) - For \( x = 9 \), \( f^{-1}(9) = 3 \) The graph of \( f^{-1}(x) = \sqrt{x} \) is a curve that starts at the origin (0,0) and rises to the right. ### 5.3 Explain why \( f^{-1}(x) \) will not be a function? The function \( f(x) = x^2 \) is not one-to-one because for any positive \( y \), there are two values of \( x \) (one positive and one negative) that yield the same \( y \). For example: - \( f(2) = 4 \) - \( f(-2) = 4 \) Thus, when we try to find the inverse, we end up with two outputs for a single input, which violates the definition of a function (which states that each input must correspond to exactly one output). Therefore, \( f^{-1}(x) \) is not a function unless we restrict the domain of \( f(x) \). ### 5.4 Explain how you will restrict the domain of \( f(x) \) to ensure that \( f^{-1}(x) \) will be a function. To ensure that \( f^{-1}(x) \) is a function, we can restrict the domain of \( f(x) = x^2 \) to non-negative values. This means we will only consider \( x \geq 0 \). By restricting the domain to \( [0, \infty) \), we ensure that each output \( y \) corresponds to exactly one input \( x \). Thus, the inverse function \( f^{-1}(x) = \sqrt{x} \) will be a valid function, as it will only take non-negative values of \( x \) and produce non-negative outputs. In summary: - The restricted domain of \( f(x) \) is \( [0, \infty) \). - The inverse function \( f^{-1}(x) = \sqrt{x} \) is now a valid function.