1. **State the problem:** Find the inverse relation of the function $$f(x) = 3x^2 - 4$$. 2. **Recall the definition:** The inverse relation swaps the roles of $$x$$ and $$y$$. If $$y = f(x)$$, then the inverse relation satisfies $$x = f(y)$$. 3. **Write the equation:** Let $$y = 3x^2 - 4$$. 4. **Swap $$x$$ and $$y$$:** To find the inverse relation, write $$x = 3y^2 - 4$$. 5. **Solve for $$y$$:** $$x = 3y^2 - 4$$ Add 4 to both sides: $$x + 4 = 3y^2$$ Divide both sides by 3: $$\frac{x + 4}{3} = y^2$$ Intermediate step showing cancellation: $$\frac{\cancel{3}y^2}{\cancel{3}} = \frac{x + 4}{3}$$ 6. **Take the square root of both sides:** $$y = \pm \sqrt{\frac{x + 4}{3}}$$ 7. **Write the inverse relation:** $$f^{-1}(x) = \pm \sqrt{\frac{x + 4}{3}}$$ **Note:** Since the original function is not one-to-one over all real numbers (because of the square), the inverse relation is not a function unless we restrict the domain of $$f$$. **Final answer:** $$f^{-1}(x) = \pm \sqrt{\frac{x + 4}{3}}$$