For \( f(x)=2 x-1 \) and \( g(x)=\frac{1}{2}(x+1) \), find \( (f \circ g)(x) \) and \( (g \circ f)(x) \). Then determine whether \( (f \circ g)(x)=(g \circ f)(x) \). What is \( (f \circ g)(x) \) ? \( (f \circ g)(x)=\square \)

To find \( (f \circ g)(x) \), we need to substitute \( g(x) \) into \( f(x) \). 1. Start with \( g(x) = \frac{1}{2}(x+1) \). 2. Now substitute this into \( f \): \[ f(g(x)) = f\left(\frac{1}{2}(x+1)\right) = 2\left(\frac{1}{2}(x+1)\right) - 1 \] 3. Simplifying gives: \[ = x + 1 - 1 = x \] Therefore, \( (f \circ g)(x) = x \). Now for \( (g \circ f)(x) \), substitute \( f(x) \) into \( g(x) \): 1. Start with \( f(x) = 2x - 1 \). 2. Now substitute this into \( g \): \[ g(f(x)) = g(2x - 1) = \frac{1}{2}((2x - 1) + 1) \] 3. Simplifying this gives: \[ = \frac{1}{2}(2x) = x \] Thus, \( (g \circ f)(x) = x \) as well. Finally, since \( (f \circ g)(x) = x \) and \( (g \circ f)(x) = x \), we can conclude that \( (f \circ g)(x) = (g \circ f)(x) \). To summarize: - \( (f \circ g)(x) = x \) - \( (g \circ f)(x) = x \) - Therefore, \( (f \circ g)(x) = (g \circ f)(x) \).