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 \( (g \circ f)(x) \) ? \( (g \circ f)(x)=\square \)
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To find \( (g \circ f)(x) \), we first need to substitute \( f(x) \) into \( g(x) \). Start with \( f(x) = 2x - 1 \). Now, plug \( f(x) \) into \( g(x) \): \[ g(f(x)) = g(2x - 1) = \frac{1}{2}((2x - 1) + 1) = \frac{1}{2}(2x) = x. \] Thus, \( (g \circ f)(x) = x \). Now let's compare \( (f \circ g)(x) \) and \( (g \circ f)(x) \): \[ (f \circ g)(x) = x \quad \text{and} \quad (g \circ f)(x) = x. \] Since both compositions yield the same result, we have \((f \circ g)(x) = (g \circ f)(x)\).