1. **State the problem:** Determine the type of discontinuity of the function $$h(x) = -3^{\frac{1}{x}+1}$$ at $$x = -1$$. 2. **Recall the types of discontinuities:** - **Removable discontinuity:** The limit exists but the function is not defined or differs at the point. - **Jump discontinuity:** The left-hand and right-hand limits exist but are not equal. - **Infinite discontinuity:** The function approaches infinity or negative infinity near the point. - **Vertical and horizontal discontinuities:** Vertical asymptotes correspond to infinite discontinuities; horizontal asymptotes describe end behavior. 3. **Analyze the function at $$x = -1$$:** Rewrite the exponent: $$\frac{1}{x} + 1 = \frac{1}{x} + \frac{x}{x} = \frac{1+x}{x}$$ At $$x = -1$$: $$\frac{1 + (-1)}{-1} = \frac{0}{-1} = 0$$ 4. **Evaluate the function at $$x = -1$$:** $$h(-1) = -3^0 = -1$$ 5. **Check the limit from the left ($$x \to -1^-$$):** For $$x$$ slightly less than $$-1$$, $$1+x < 0$$, so the exponent $$\frac{1+x}{x}$$ is negative and approaches 0 from the negative side. Thus, $$3^{\text{small negative}} \to 3^0 = 1$$ So, $$h(x) \to -1$$ from the left. 6. **Check the limit from the right ($$x \to -1^+$$):** For $$x$$ slightly greater than $$-1$$, $$1+x > 0$$, so the exponent $$\frac{1+x}{x}$$ is positive and approaches 0 from the positive side. Thus, $$3^{\text{small positive}} \to 3^0 = 1$$ So, $$h(x) \to -1$$ from the right. 7. **Conclusion:** Since $$\lim_{x \to -1^-} h(x) = \lim_{x \to -1^+} h(x) = h(-1) = -1$$, the function is continuous at $$x = -1$$. Therefore, there is **no discontinuity** at $$x = -1$$. **Note:** The problem asks to determine the type of discontinuity, but the function is continuous at $$x = -1$$, so none of the discontinuity types apply here.