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)

Let's solve the problem step by step. ### 6.1 Calculate the value of \( a \). We know that the point \( (8, 4) \) lies on the graph of \( g \), which is defined by the equation \( g(x) = \sqrt{a x} \). This means that when \( x = 8 \), \( g(8) = 4 \). Substituting into the equation: \[ g(8) = \sqrt{a \cdot 8} = 4 \] Now, squaring both sides to eliminate the square root gives: \[ a \cdot 8 = 16 \] Now, solving for \( a \): \[ a = \frac{16}{8} = 2 \] ### 6.2 If \( g(x) > 0 \), for what values of \( x \) will \( g \) be defined? The function \( g(x) = \sqrt{a x} \) is defined when the expression inside the square root is non-negative: \[ a x \geq 0 \] Since we found \( a = 2 \): \[ 2x \geq 0 \implies x \geq 0 \] Thus, \( g \) is defined for \( x \geq 0 \). ### 6.3 Determine the range of \( g \). The function \( g(x) = \sqrt{2x} \) takes non-negative values since the square root function outputs non-negative results. As \( x \) approaches infinity, \( g(x) \) also approaches infinity. When \( x = 0 \), \( g(0) = 0 \). Thus, the range of \( g \) is: \[ [0, \infty) \] ### 6.4 Write down the equation of \( g^{-1} \), the inverse of \( g \), in the form \( y = \ldots \). To find the inverse, we start with the equation: \[ y = g(x) = \sqrt{2x} \] Now, we solve for \( x \): 1. Square both sides: \[ y^2 = 2x \] 2. Solve for \( x \): \[ x = \frac{y^2}{2} \] Thus, the inverse function is: \[ g^{-1}(y) = \frac{y^2}{2} \] In the form \( y = \ldots \), we can write: \[ y = g^{-1}(x) = \frac{x^2}{2} \] ### 6.5 Determine ALGEBRAICALLY the point(s) of intersection of \( h \) and \( g \). The function \( h(x) = x - 4 \). To find the points of intersection, we set \( g(x) \) equal to \( h(x) \): \[ \sqrt{2x} = x - 4 \] Now, squaring both sides: \[ 2x = (x - 4)^2 \] Expanding the right side: \[ 2x = x^2 - 8x + 16 \] Rearranging gives: \[ 0 = x^2 - 10x + 16 \] Now, we can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 1, b = -10, c = 16 \): \[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} \] \[ x = \frac{10 \pm \sqrt{100 - 64}}{2} \] \[ x = \frac{10 \pm \sqrt{36}}{2} \] \[ x = \frac{10 \pm 6}{2} \] Calculating the two possible values: 1. \( x = \frac{16}{2} = 8 \) 2. \( x = \frac{4}{2} = 2 \) Thus, the points of intersection are at \( x = 2 \) and \( x = 8 \). ### Determine the values of \( x \) for which \( g(x) > h(x) \). To find where \( g(x) > h(x) \), we set up the inequality: \[ \sqrt{2x} > x - 4 \] Squaring both sides (noting that both sides must be non-negative): \[ 2x > (x - 4)^2 \] This leads to the same quadratic equation we solved earlier: \[ 0 > x^2 - 10x + 16 \] Factoring gives: \[ 0 > (x - 2)(x - 8) \] The solution to this inequality is found by testing intervals. The critical points are \( x = 2 \) and \( x = 8 \). The inequality holds between these points: Thus, \( g(x) > h(x) \) for: \[ 2 < x < 8 \] ### Summary of Answers: - 6.1: \( a = 2 \) - 6.2: \( x \geq 0 \) - 6.3: Range of \( g \) is \( [0, \infty) \) - 6.4: \( g^{-1}(x) = \frac{x^2}{2} \) - 6.5: Points of intersection at \( x = 2 \) and \( x = 8 \); \( g(x) > h(x) \) for \( 2 < x < 8 \).