1. **State the problem:** Given functions $f(x) = x^3 - 2$ and $g(x) = \frac{5x + 3}{1 - 2x}$, find the composition $(g \circ f)(x)$ and the inverse function $g^{-1}(x)$. 2. **Find $(g \circ f)(x)$:** This means $g(f(x))$. Substitute $f(x)$ into $g$: $$g(f(x)) = \frac{5(f(x)) + 3}{1 - 2(f(x))} = \frac{5(x^3 - 2) + 3}{1 - 2(x^3 - 2)}$$ 3. **Simplify numerator and denominator:** Numerator: $$5(x^3 - 2) + 3 = 5x^3 - 10 + 3 = 5x^3 - 7$$ Denominator: $$1 - 2(x^3 - 2) = 1 - 2x^3 + 4 = 5 - 2x^3$$ So, $$g(f(x)) = \frac{5x^3 - 7}{5 - 2x^3}$$ 4. **Find $g^{-1}(x)$:** To find the inverse, start with $$y = \frac{5x + 3}{1 - 2x}$$ Swap $x$ and $y$: $$x = \frac{5y + 3}{1 - 2y}$$ 5. **Solve for $y$:** Multiply both sides by denominator: $$x(1 - 2y) = 5y + 3$$ Distribute: $$x - 2xy = 5y + 3$$ Group $y$ terms on one side: $$-2xy - 5y = 3 - x$$ Factor out $y$: $$y(-2x - 5) = 3 - x$$ 6. **Divide both sides by $-2x - 5$:** $$y = \frac{3 - x}{-2x - 5}$$ Rewrite denominator: $$y = \frac{3 - x}{-(2x + 5)} = -\frac{3 - x}{2x + 5} = \frac{x - 3}{2x + 5}$$ 7. **Final answers:** $$(g \circ f)(x) = \frac{5x^3 - 7}{5 - 2x^3}$$ $$g^{-1}(x) = \frac{x - 3}{2x + 5}$$