1. **State the problem:** Show that for the functions $f(x) = 7x - 2$ and $g(x) = \frac{1}{7}(x + 2)$, the compositions satisfy $(f \circ g)(x) = (g \circ f)(x) = x$. 2. **Recall the definition of composition:** - $(f \circ g)(x) = f(g(x))$ - $(g \circ f)(x) = g(f(x))$ 3. **Calculate $(f \circ g)(x)$:** $$ (f \circ g)(x) = f\left(g(x)\right) = f\left(\frac{1}{7}(x + 2)\right) = 7 \cdot \frac{1}{7}(x + 2) - 2 $$ Simplify the multiplication: $$ 7 \cdot \frac{1}{7}(x + 2) = \cancel{7} \cdot \frac{1}{\cancel{7}}(x + 2) = x + 2 $$ So, $$ (f \circ g)(x) = x + 2 - 2 = x $$ 4. **Calculate $(g \circ f)(x)$:** $$ (g \circ f)(x) = g\left(f(x)\right) = g(7x - 2) = \frac{1}{7}((7x - 2) + 2) $$ Simplify inside the parentheses: $$ (7x - 2) + 2 = 7x $$ So, $$ (g \circ f)(x) = \frac{1}{7} \cdot 7x = \cancel{\frac{1}{7}} \cdot \cancel{7} x = x $$ 5. **Conclusion:** Both compositions return $x$, so $(f \circ g)(x) = (g \circ f)(x) = x$. This shows that $f$ and $g$ are inverse functions of each other.