1. **Problem Statement:** Find the inverse of the function $f_1(x) = -3x^2 + 4$. 2. **Recall the formula and rules:** To find the inverse function $f_1^{-1}(x)$, we swap $x$ and $y$ in the equation and solve for $y$. 3. **Step 1: Write the function as an equation:** $$y = -3x^2 + 4$$ 4. **Step 2: Swap $x$ and $y$:** $$x = -3y^2 + 4$$ 5. **Step 3: Solve for $y$:** $$x - 4 = -3y^2$$ $$\frac{x - 4}{-3} = y^2$$ $$y^2 = \frac{4 - x}{3}$$ 6. **Step 4: Take the square root of both sides:** $$y = \pm \sqrt{\frac{4 - x}{3}}$$ 7. **Step 5: Determine the domain and range for the inverse:** Since $f_1(x)$ is a quadratic opening downward, it is not one-to-one over all real numbers. To have an inverse, restrict the domain of $f_1$ to where it is one-to-one (e.g., $x \geq 0$ or $x \leq 0$). Assuming $x \geq 0$, the inverse is: $$f_1^{-1}(x) = -\sqrt{\frac{4 - x}{3}}$$ If $x \leq 0$, then: $$f_1^{-1}(x) = +\sqrt{\frac{4 - x}{3}}$$ 8. **Final answer:** $$f_1^{-1}(x) = \pm \sqrt{\frac{4 - x}{3}}$$ with domain $x \leq 4$ (since $4 - x \geq 0$) and range depending on the chosen branch.