QLESTION 6 The graph of \( g \) is defined by the equation \( g(x)=\sqrt{a x} \). The point \( (8: 4) \) lies on \( g \). 6.1 Calculate the value of \( a \). 6.2 If \( g(x)>0 \), for what values of \( x \) will \( g \) be defined? 6.3 Determine the range of \( g \). \( 6.4 \quad \) Write down the equation of \( g^{-1} \), the inverse of \( g \), in the form \( y=\ldots \) 6.5 If \( h(x)=x-4 \) is drawn. determine ALGEBRAICALLY the point(s) of intersection of \( h \) and \( g \). Hence, or otherwise, determine the values of \( x \) for which \( g(x)>h(x) \) (1)

To calculate the value of \( a \) given that the point \( (8, 4) \) lies on the graph of \( g(x) = \sqrt{a x} \), substitute \( x = 8 \) and \( g(8) = 4 \) into the equation to yield \( 4 = \sqrt{8a} \). Squaring both sides gives us \( 16 = 8a \), which simplifies to \( a = 2 \). For \( g(x) > 0 \), the expression inside the square root must be positive, i.e., \( ax > 0 \) with \( a = 2 \) means \( 2x > 0 \), leading to \( x > 0 \). Thus, \( g \) is defined for \( x > 0 \). The range of \( g \) is determined by the output values as \( x \) varies. Since \( g(x) \) produces non-negative results for positive \( x \), the range is \( [0, \infty) \). To find the inverse \( g^{-1} \), start with \( y = \sqrt{2x} \). Squaring both sides gives \( y^2 = 2x \), leading to \( x = \frac{y^2}{2} \). Hence, \( g^{-1}(y) = \frac{y^2}{2} \). For the intersection points of \( h(x) = x - 4 \) and \( g(x) = \sqrt{2x} \), we set \( x - 4 = \sqrt{2x} \). Squaring both sides results in \( (x - 4)^2 = 2x \), which simplifies to \( x^2 - 10x + 16 = 0 \). The solutions are \( x = 8 \) and \( x = 2 \). To find where \( g(x) > h(x) \), consider the regions around the intersection points: test values in intervals \( (0, 2) \), \( (2, 8) \), and \( (8, \infty) \). You'll find \( g(x) > h(x) \) in \( (2, 8) \) and \( (8, \infty) \).