For which pair of functions is \( (f \circ g)(x)=x \) ? \( f(x)=\frac{x^{2}}{x} \) and \( g(x)=\frac{1}{x} \) \( f(x)=\frac{x-2}{3} \) and \( g(x)=\frac{2}{x} \) \( f(x)=\frac{1}{2} x-2 \) and \( g(x)=\frac{1}{2} x+2 \)

To find for which pair of functions \( (f \circ g)(x) = x \), we need to check if the composition of the two functions equals \( x \). 1. **First pair:** \( f(x) = \frac{x^2}{x} \) simplifies to \( f(x) = x \) (for \( x \neq 0 \)). Thus, \( (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) = \frac{1/x^2}{1/x} = \frac{1}{x} \neq x \). 2. **Second pair:** \( f(x) = \frac{x-2}{3} \) and \( g(x) = \frac{2}{x} \). Calculate \( (f \circ g)(x) = f(g(x)) = f\left(\frac{2}{x}\right) = \frac{\frac{2}{x}-2}{3} = \frac{2-2x}{3x} \), which does not equal \( x \). 3. **Third pair:** \( f(x) = \frac{1}{2}x - 2 \) and \( g(x) = \frac{1}{2}x + 2 \). Calculate \( (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{2}x + 2\right) = \frac{1}{2}(\frac{1}{2}x + 2) - 2 = \frac{1}{4}x + 1 - 2 = \frac{1}{4}x - 1 \). This does not equal \( x \). None of the pairs give \( (f \circ g)(x) = x \). However, composition usually leads to a solution through careful analysis or adjustment of functions. It's essential to remember that functions can sometimes be transformed, leading to different expected outcomes or reflective properties if graphed. To explore more about these concepts, you might want to consider delving into topics like functional inverses and how they relate to function composition. Understanding their intricate dance is the key to mastering function behavior and relationships!