Inverse Functions D6Ccc5

1. **State the problem:** We need to determine which pair of functions are inverses of each other. 2. **Recall the definition of inverse functions:** Two functions $f$ and $g$ are inverses if and only if $f(g(x)) = x$ and $g(f(x)) = x$ for all $x$ in the domains. 3. **Check each pair:** **A.** $$f(g(x)) = 10 \cdot \frac{x+5}{10} - 5 = (x+5) - 5 = x$$ $$g(f(x)) = \frac{10x - 5 + 5}{10} = \frac{10x}{10} = x$$ So, pair A are inverses. **B.** $$f(g(x)) = \frac{8x - 3}{8} + 3 = x - \frac{3}{8} + 3 \neq x$$ $$g(f(x)) = 8 \left(\frac{x}{8} + 3\right) - 3 = x + 24 - 3 = x + 21 \neq x$$ So, pair B are not inverses. **C.** $$f(g(x)) = \frac{4}{\frac{x+3}{4}} - 3 = \frac{4 \cdot 4}{x+3} - 3 = \frac{16}{x+3} - 3 \neq x$$ $$g(f(x)) = \frac{\frac{4}{x} - 3 + 3}{4} = \frac{4/x}{4} = \frac{1}{x} \neq x$$ So, pair C are not inverses. **D.** $$f(g(x)) = \sqrt[3]{6 \cdot \left(\frac{x}{6}\right)^3} = \sqrt[3]{6 \cdot \frac{x^3}{216}} = \sqrt[3]{\frac{x^3}{36}} = \frac{x}{\sqrt[3]{36}} \neq x$$ $$g(f(x)) = \left(\frac{\sqrt[3]{6x}}{6}\right)^3 = \left(\frac{(6x)^{1/3}}{6}\right)^3 = \frac{6x}{216} = \frac{x}{36} \neq x$$ So, pair D are not inverses. 4. **Conclusion:** Only pair A are inverses of each other. **Final answer:** Pair A