QUESTION 3 Giver: \( f(x)=\frac{2}{x-3}-1 \) and \( g(x)=x-3 \) 3.1 Write down the donain of \( f \). 3.2 Calculate the \( x \)-irtercept of \( f \). (1) 3.3 Draw the graplas of \( f \) and \( g \) on the surse syatem of axes. Clearly show all the intercepts with the axes os well as the nsyroptotes. (2) (5) 3.4 Calculate the volue(s) of \( x \) for which \( f(x) \geq g(x) \) if \( x \geq 3 \). (f) [12]

The function \( f(x) = \frac{2}{x-3} - 1 \) has a domain that excludes the value where the denominator becomes zero, which is \( x = 3 \). Therefore, the domain of \( f \) is all real numbers except \( 3 \), or in interval notation: \( (-\infty, 3) \cup (3, \infty) \). To find the x-intercept of \( f \), you set \( f(x) = 0 \) and solve for \( x \). This leads to the equation \( \frac{2}{x-3} - 1 = 0 \), simplifying to \( \frac{2}{x-3} = 1 \). Solving gives \( x - 3 = 2 \) or \( x = 5 \). Thus, the x-intercept of \( f \) is \( (5, 0) \). To graph \( f \) and \( g(x) = x - 3 \), start with the basic characteristics: \( f \) has vertical asymptote at \( x = 3 \) and approaches horizontal asymptote \( y = -1 \). The function \( g(x) \) is a straight line passing through \( (3, 0) \) with a slope of 1. Mark the intercepts, asymptotes, and plot their intersections for a clear visualization! For the inequality \( f(x) \geq g(x) \) with \( x \geq 3 \), compare the two functions in the relevant intervals. This requires solving \( \frac{2}{x-3} - 1 \geq x - 3 \). Simplifying leads to \( \frac{2}{x-3} \geq x - 2 \), which when manipulated and solved will reveal the regions where \( f \) is above \( g \). Happy solving! 🌟