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 \( (f \circ g)(x) \) ? \( (f \circ g)(x)=x \) What is \( (g \circ f)(x) \) ? \( (g \circ f)(x)=\square \)

Let's solve the problem step by step. ### Step 1. Compute \( (f \circ g)(x) \) By definition, \[ (f \circ g)(x) = f(g(x)) \] Given: \[ f(x) = 2x-1 \quad \text{and} \quad g(x) = \frac{1}{2}(x+1) \] Substitute \( g(x) \) into \( f(x) \): \[ f(g(x)) = 2\left(\frac{1}{2}(x+1)\right) - 1 \] Simplify the expression: \[ 2\left(\frac{1}{2}(x+1)\right) = x+1 \] \[ x+1 - 1 = x \] Thus: \[ (f \circ g)(x) = x \] ### Step 2. Compute \( (g \circ f)(x) \) By definition, \[ (g \circ f)(x) = g(f(x)) \] Substitute \( f(x) \) into \( g(x) \): \[ g(f(x)) = \frac{1}{2}\left((2x-1)+1\right) \] Simplify the expression within the parentheses: \[ (2x-1)+1 = 2x \] Now compute: \[ \frac{1}{2}(2x) = x \] Thus: \[ (g \circ f)(x) = x \] ### Step 3. Compare the Two Compositions We found: \[ (f \circ g)(x) = x \quad \text{and} \quad (g \circ f)(x) = x \] Therefore, they are equal: \[ (f \circ g)(x) = (g \circ f)(x) \] ### Final Answer - \( (f \circ g)(x) = x \) - \( (g \circ f)(x) = x \) Since both compositions result in \( x \), they are indeed equal.