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 \( f(x) \).
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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 \( [0, \infty) \).