1. **State the problem:** We need to analyze the function $$f(x) = \frac{1}{(x+1)^2} + 3$$ at $$x = -1$$ and determine the type of discontinuity. 2. **Recall the definition:** An infinite (essential) discontinuity occurs when the function approaches infinity or negative infinity near a point. 3. **Evaluate the function near $$x = -1$$:** - The denominator is $$(x+1)^2$$ which becomes zero at $$x = -1$$. - Since the denominator is squared, it is always positive except at $$x = -1$$ where it is zero. 4. **Check the behavior from the left and right:** - As $$x \to -1^-$$, $$(x+1)^2 \to 0^+$$, so $$\frac{1}{(x+1)^2} \to +\infty$$. - As $$x \to -1^+$$, similarly, $$(x+1)^2 \to 0^+$$, so $$\frac{1}{(x+1)^2} \to +\infty$$. 5. **Add the constant 3:** - Since $$\frac{1}{(x+1)^2} \to +\infty$$, adding 3 does not change the infinite behavior. 6. **Conclusion:** - The function has an infinite discontinuity at $$x = -1$$ because the function tends to infinity from both sides. **Final answer:** The function $$f(x) = \frac{1}{(x+1)^2} + 3$$ has an infinite essential discontinuity at $$x = -1$$.