1. **State the problem:** Evaluate the improper integral $$\int_{-\infty}^{0} \frac{1}{\sqrt[4]{4-x}} \, dx$$ and determine if it converges or diverges. 2. **Rewrite the integral using a limit:** Since the lower limit is $-\infty$, write $$\int_{-\infty}^{0} \frac{1}{\sqrt[4]{4-x}} \, dx = \lim_{b \to -\infty} \int_{b}^{0} \frac{1}{\sqrt[4]{4-x}} \, dx.$$ Note the limit is as $b \to -\infty$ (not $\infty$). 3. **Substitution:** Let $$u = 4 - x \implies du = -dx \implies dx = -du.$$ When $x = b$, $u = 4 - b$; when $x=0$, $u=4$. 4. **Change the integral in terms of $u$:** $$\int_{b}^{0} \frac{1}{\sqrt[4]{4-x}} \, dx = \int_{u=4-b}^{4} u^{-1/4} (-du) = \int_{4}^{4-b} u^{-1/4} \, du.$$ We reversed limits because of the negative sign. 5. **Integrate:** $$\int u^{-1/4} \, du = \frac{u^{3/4}}{3/4} = \frac{4}{3} u^{3/4}.$$ So, $$\int_{4}^{4-b} u^{-1/4} \, du = \left. \frac{4}{3} u^{3/4} \right|_{4}^{4-b} = \frac{4}{3} \left( (4-b)^{3/4} - 4^{3/4} \right).$$ 6. **Evaluate the limit:** As $b \to -\infty$, $4 - b \to \infty$, so $$\lim_{b \to -\infty} (4-b)^{3/4} = \infty.$$ Therefore, $$\lim_{b \to -\infty} \int_{b}^{0} \frac{1}{\sqrt[4]{4-x}} \, dx = \lim_{b \to -\infty} \frac{4}{3} \left( (4-b)^{3/4} - 4^{3/4} \right) = \infty.$$ 7. **Conclusion:** The integral diverges to infinity. **Final answer:** $$\int_{-\infty}^{0} \frac{1}{\sqrt[4]{4-x}} \, dx = \infty.$$