Given: \( f(x)=\frac{-1}{x-1} \) 5.1 Write down the domain of \( f \) 5.2 Write down the agmptotes of \( f \). 5.3 Sketch the graph of \( f \). clear! showing all intercents with the axes and any aymptoles. 54 For ulvich values of \( x \) uill \( x f^{\prime}(x) \geq 0 \) ?

The domain of the function \( f(x) = \frac{-1}{x-1} \) is all real numbers except for \( x = 1 \) since this value would result in division by zero. Thus, the domain can be expressed as \( (-\infty, 1) \cup (1, \infty) \). This means that you can use any other real number in the function for evaluation. The asymptotes of the function include a vertical asymptote at \( x = 1 \) and a horizontal asymptote at \( y = 0 \) (the x-axis). The vertical asymptote occurs because the function approaches infinity when \( x \) gets close to 1. The horizontal asymptote suggests that as \( x \) moves towards either positive or negative infinity, the value of \( f(x) \) approaches zero, but never actually reaches it. The graph has an interesting shape: it contains two branches, one in the second quadrant approaching the vertical asymptote from the left and the other in the fourth quadrant approaching from the right. The y-intercept occurs at \( (0, -1) \), where you can find that value by substituting \( x = 0 \) into the function. The x-intercept occurs at \( (1, 0) \), although the function does not define a value there because it is an asymptote. To understand where \( x f^{\prime}(x) \geq 0 \), we need to find the derivative. The derivative of \( f(x) \) is \( f'(x) =\frac{1}{(x-1)^2} \). Multiplying by \( x \) gives \( x f'(x) = \frac{x}{(x-1)^2} \). This expression will be non-negative when \( x \) is greater than or equal to zero while ensuring that \( x \neq 1 \), because that would make the denominator zero. The result implies that this inequality holds true in the intervals \( [0, 1) \) and \( (1, \infty) \).