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be a furnction. 4 Given: \( f(x)=\frac{1}{2} x^{2} \) and \( g(x)=2 x+4 \) 4.1 Is \( f(x) \) a function? Give a reason for your answer. 4.2 Write down the domain of \( f(x) \) for \( f^{-1}(x) \) to be a function 4.3 Write down the equation of \( g^{-1}(x) \) in the form \( y=\ldots \) 4.4 Write down the range of \( f(x) \).

Ask by Guzman Grant. in South Africa
Jan 23,2025

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Answer

**4.1** Yes, \( f(x) = \frac{1}{2} x^{2} \) is a function because each input \( x \) has exactly one output. **4.2** To make \( f^{-1}(x) \) a function, the domain of \( f(x) \) should be restricted to \( x \geq 0 \). **4.3** The inverse function \( g^{-1}(x) \) is \( y = \frac{x - 4}{2} \). **4.4** The range of \( f(x) \) is all real numbers \( y \) where \( y \geq 0 \).

Solution

Certainly! Let's address each part of the problem step by step. ### **Given:** - \( f(x) = \frac{1}{2} x^{2} \) - \( g(x) = 2x + 4 \) --- ### **4.1 Is \( f(x) \) a function?** **Answer:** Yes, \( f(x) = \frac{1}{2} x^{2} \) is a function. **Reason:** For every input value \( x \) in the domain, there is exactly one output value \( f(x) \). This satisfies the definition of a function, where each input is associated with a single output. --- ### **4.2 Domain of \( f(x) \) for \( f^{-1}(x) \) to be a function** **Answer:** The domain of \( f(x) \) must be restricted to \( x \geq 0 \) (i.e., all non-negative real numbers). **Explanation:** The function \( f(x) = \frac{1}{2} x^{2} \) is not one-to-one over all real numbers because both \( x \) and \( -x \) yield the same output. To ensure that the inverse \( f^{-1}(x) \) is also a function (i.e., passes the horizontal line test), we need to restrict the domain of \( f(x) \) to either \( x \geq 0 \) or \( x \leq 0 \). Typically, the domain is restricted to \( x \geq 0 \). --- ### **4.3 Equation of \( g^{-1}(x) \)** **Answer:** The inverse function \( g^{-1}(x) \) is given by: \[ y = \frac{x - 4}{2} \] **Derivation:** 1. Start with \( y = 2x + 4 \). 2. Swap \( x \) and \( y \): \( x = 2y + 4 \). 3. Solve for \( y \): \[ x - 4 = 2y \\ y = \frac{x - 4}{2} \] So, \( g^{-1}(x) = \frac{x - 4}{2} \). --- ### **4.4 Range of \( f(x) \)** **Answer:** The range of \( f(x) = \frac{1}{2} x^{2} \) is all real numbers \( y \) such that \( y \geq 0 \). **Explanation:** Since \( f(x) \) is a quadratic function that opens upwards and its minimum value occurs at \( x = 0 \), where \( f(0) = 0 \), the function outputs are all non-negative real numbers. Therefore, the range is \( y \geq 0 \). --- If you have any further questions or need more clarification on any part, feel free to ask!

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The function \( f(x) = \frac{1}{2}x^2 \) is indeed a function, as it assigns exactly one output to each input \( x \). Since it's a quadratic function, it passes the vertical line test; any vertical line will intersect the graph at most once! For \( f^{-1}(x) \) to be a function, the domain must be restricted to non-negative values of \( f(x) \), as the parabola opens upwards. Thus, the domain of \( f(x) \) for it to have an inverse would be \( x \geq 0 \). To find the inverse of \( g(x) = 2x + 4 \), we switch \( x \) and \( y \) and solve for \( y \): 1. Start with \( x = 2y + 4 \). 2. Rearranging gives \( y = \frac{x - 4}{2} \). So the equation of the inverse \( g^{-1}(x) \) is \( y = \frac{x - 4}{2} \). The range of \( f(x) = \frac{1}{2}x^2 \) is all non-negative real numbers, as the lowest value occurs at \( f(0) = 0 \) and it increases without bound. Thus, the range is \( f(x) \geq 0 \) or \( [0, \infty) \).

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