1. **State the problem:** We are given two functions: $$f(x) = \frac{x^3 + 2x^2 - 4x - 8}{x^2 - 4} \quad \text{and} \quad g(x) = \frac{x^3 + 6x^2 + 12x + 8}{x^2 + 4x + 4}$$ We need to analyze their equality, values at specific points, and limits. 2. **Factor the denominators:** $$x^2 - 4 = (x-2)(x+2)$$ $$x^2 + 4x + 4 = (x+2)^2$$ 3. **Factor the numerators:** For $f(x)$ numerator: $$x^3 + 2x^2 - 4x - 8$$ Group terms: $$= (x^3 + 2x^2) - (4x + 8) = x^2(x+2) - 4(x+2)$$ Factor out $(x+2)$: $$= (x+2)(x^2 - 4)$$ Recall $x^2 - 4 = (x-2)(x+2)$, so numerator is: $$ (x+2)(x-2)(x+2) = (x+2)^2 (x-2)$$ For $g(x)$ numerator: $$x^3 + 6x^2 + 12x + 8$$ Try factoring by grouping: $$= (x^3 + 6x^2) + (12x + 8) = x^2(x+6) + 4(3x+2)$$ This doesn't factor nicely by grouping, try synthetic division or recognize it as a cube: Note that $(x+2)^3 = x^3 + 6x^2 + 12x + 8$ So numerator of $g(x)$ is: $$ (x+2)^3$$ 4. **Rewrite functions:** $$f(x) = \frac{(x+2)^2 (x-2)}{(x-2)(x+2)}$$ $$g(x) = \frac{(x+2)^3}{(x+2)^2}$$ 5. **Simplify by canceling common factors:** For $f(x)$: $$f(x) = \frac{(x+2)^2 \cancel{(x-2)}}{\cancel{(x-2)} (x+2)} = \frac{(x+2)^2}{x+2}$$ Intermediate step with cancellation: $$f(x) = \frac{(x+2)^2 \cancel{(x-2)}}{\cancel{(x-2)} (x+2)}$$ Simplify numerator and denominator: $$= x+2, \quad x \neq 2, -2$$ For $g(x)$: $$g(x) = \frac{(x+2)^3}{(x+2)^2} = x+2, \quad x \neq -2$$ 6. **Domain considerations:** - $f(x)$ is undefined at $x=2$ and $x=-2$ because denominator zeroes. - $g(x)$ is undefined at $x=-2$. 7. **Evaluate $f(-2)$ and $g(-2)$:** - $f(-2)$ denominator zero, so undefined. - $g(-2)$ denominator zero, so undefined. So $f(-2) \neq g(-2)$ because both undefined. 8. **Evaluate $f(2)$ and $g(2)$:** - $f(2)$ denominator zero, undefined. - $g(2) = 2 + 2 = 4$ So $f(2)$ undefined, $g(2) = 4$. 9. **Limits at $x=2$:** Since $f(x) = x+2$ for $x \neq 2$, limit of $f(x)$ as $x \to 2$ is: $$\lim_{x \to 2} f(x) = 2 + 2 = 4$$ Similarly, $g(2) = 4$ and $g(x) = x+2$ for $x \neq -2$, so limit at 2 is also 4. 10. **Conclusion:** - $f(x)$ and $g(x)$ are equal everywhere except at $x=2$ and $x=-2$ where they are undefined. - Their limits at $x=2$ are equal. - $f(-2)$ and $g(-2)$ are both undefined, so $f(-2) \neq g(-2)$ is true. **Correct statement:** "While $f(x)$ and $g(x)$ appear to be different functions, in fact they have the same values everywhere except $x=2$. That's why the limit of $f(x)$ and $g(x)$ are different at $x=2$." is true. Also, "$f(2) \neq g(2)$. But $f(x) = g(x)$ are equal everywhere else and that's why their limits are the same everywhere." is true. Therefore, there are exactly two correct answers (not counting the answer choices that say so). --- **Final answers:** - $f(x) = g(x) = x+2$ for all $x \neq 2, -2$ - $f(2)$ undefined, $g(2) = 4$ - $f(-2)$ undefined, $g(-2)$ undefined - Limits at $x=2$ are equal to 4 - Two correct statements from the options