1. **State the problem:** Determine if the given pairs of functions are inverses of each other. 2. **Recall the definition:** 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:** **Pair 1:** $$g(x) = 4 - \frac{3}{2}x, \quad f(x) = \frac{1}{2}x + \frac{3}{2}$$ Calculate $f(g(x))$: $$f(g(x)) = \frac{1}{2}\left(4 - \frac{3}{2}x\right) + \frac{3}{2} = 2 - \frac{3}{4}x + \frac{3}{2} = \frac{7}{2} - \frac{3}{4}x$$ Since $f(g(x)) \neq x$, they are not inverses. **Pair 2:** $$g(n) = \frac{-12 - 2n}{3}, \quad f(n) = \frac{-5 + 6n}{5}$$ Calculate $f(g(n))$: $$f\left(\frac{-12 - 2n}{3}\right) = \frac{-5 + 6\left(\frac{-12 - 2n}{3}\right)}{5} = \frac{-5 + 2(-12 - 2n)}{5} = \frac{-5 - 24 - 4n}{5} = \frac{-29 - 4n}{5}$$ Since $f(g(n)) \neq n$, they are not inverses. **Pair 3:** $$f(n) = \frac{-16 + n}{4}, \quad g(n) = 4n + 16$$ Calculate $f(g(n))$: $$f(4n + 16) = \frac{-16 + (4n + 16)}{4} = \frac{4n}{4} = n$$ Calculate $g(f(n))$: $$g\left(\frac{-16 + n}{4}\right) = 4\left(\frac{-16 + n}{4}\right) + 16 = (-16 + n) + 16 = n$$ Since both compositions equal $n$, they are inverses. **Pair 4:** $$f(x) = -\frac{4}{7}x - \frac{16}{7}, \quad g(x) = \frac{3}{2}x - \frac{3}{2}$$ Calculate $f(g(x))$: $$f\left(\frac{3}{2}x - \frac{3}{2}\right) = -\frac{4}{7}\left(\frac{3}{2}x - \frac{3}{2}\right) - \frac{16}{7} = -\frac{6}{7}x + \frac{6}{7} - \frac{16}{7} = -\frac{6}{7}x - \frac{10}{7}$$ Since $f(g(x)) \neq x$, they are not inverses. **Pair 5:** $$f(n) = -(n + 1)^3, \quad g(n) = 3 + n^3$$ Calculate $f(g(n))$: $$f(3 + n^3) = -\left(3 + n^3 + 1\right)^3 = -\left(n^3 + 4\right)^3$$ This is not equal to $n$, so not inverses. **Pair 6:** $$f(n) = 2(n - 2)^3, \quad g(n) = \frac{4 + \sqrt[3]{4n}}{2}$$ Calculate $f(g(n))$: $$f\left(\frac{4 + \sqrt[3]{4n}}{2}\right) = 2\left(\frac{4 + \sqrt[3]{4n}}{2} - 2\right)^3 = 2\left(\frac{4 + \sqrt[3]{4n} - 4}{2}\right)^3 = 2\left(\frac{\sqrt[3]{4n}}{2}\right)^3 = 2\frac{4n}{8} = n$$ Calculate $g(f(n))$: $$g(2(n - 2)^3) = \frac{4 + \sqrt[3]{4 \cdot 2(n - 2)^3}}{2} = \frac{4 + \sqrt[3]{8(n - 2)^3}}{2} = \frac{4 + 2(n - 2)}{2} = n$$ They are inverses. **Pair 7:** $$f(x) = \frac{4}{-x - 2} + 2, \quad h(x) = -\frac{1}{x + 3}$$ Calculate $f(h(x))$: $$f\left(-\frac{1}{x + 3}\right) = \frac{4}{-\left(-\frac{1}{x + 3}\right) - 2} + 2 = \frac{4}{\frac{1}{x + 3} - 2} + 2 = \frac{4}{\frac{1 - 2(x + 3)}{x + 3}} + 2 = \frac{4(x + 3)}{1 - 2x - 6} + 2 = \frac{4(x + 3)}{-2x - 5} + 2$$ Not equal to $x$, so not inverses. **Pair 8:** $$g(x) = -\frac{2}{x} - 1, \quad f(x) = -\frac{2}{x + 1}$$ Calculate $g(f(x))$: $$g\left(-\frac{2}{x + 1}\right) = -\frac{2}{-\frac{2}{x + 1}} - 1 = -\frac{2}{-\frac{2}{x + 1}} - 1 = -\left(-\frac{x + 1}{1}\right) - 1 = (x + 1) - 1 = x$$ Calculate $f(g(x))$: $$f\left(-\frac{2}{x} - 1\right) = -\frac{2}{-\frac{2}{x} - 1 + 1} = -\frac{2}{-\frac{2}{x}} = -\left(-\frac{x}{1}\right) = x$$ They are inverses. 4. **Find inverses for given functions:** **9)** $h(x) = \sqrt[3]{x} - 3$ Set $y = \sqrt[3]{x} - 3$, solve for $x$: $$y + 3 = \sqrt[3]{x} \implies (y + 3)^3 = x$$ Inverse: $$h^{-1}(x) = (x + 3)^3$$ **10)** $g(x) = \frac{1}{x} - 2$ Set $y = \frac{1}{x} - 2$, solve for $x$: $$y + 2 = \frac{1}{x} \implies x = \frac{1}{y + 2}$$ Inverse: $$g^{-1}(x) = \frac{1}{x + 2}$$ **11)** $h(x) = 2x^3 + 3$ Set $y = 2x^3 + 3$, solve for $x$: $$y - 3 = 2x^3 \implies x^3 = \frac{y - 3}{2} \implies x = \sqrt[3]{\frac{y - 3}{2}}$$ Inverse: $$h^{-1}(x) = \sqrt[3]{\frac{x - 3}{2}}$$ **12)** $g(x) = -4x + 1$ Set $y = -4x + 1$, solve for $x$: $$y - 1 = -4x \implies x = -\frac{y - 1}{4} = \frac{1 - y}{4}$$ Inverse: $$g^{-1}(x) = \frac{1 - x}{4}$$