1. **State the problem:** We have the function $f(x) = x^2 - x - 2$ and need to: (i) Simplify $\frac{f(x)}{f(x+3)}$ to $\frac{x-2}{x+4}$ when $f(x+3) \neq 0$. (ii) Find the range of $x$ such that $\frac{f(x)}{f(x+3)} < 2$ with $x \neq -4$. 2. **Recall the formula and factorization:** The quadratic $f(x) = x^2 - x - 2$ factors as: $$f(x) = (x - 2)(x + 1)$$ Similarly, for $f(x+3)$: $$f(x+3) = (x+3)^2 - (x+3) - 2$$ 3. **Simplify $f(x+3)$:** $$f(x+3) = (x+3)^2 - (x+3) - 2 = (x^2 + 6x + 9) - x - 3 - 2 = x^2 + 5x + 4$$ Factor $f(x+3)$: $$x^2 + 5x + 4 = (x + 4)(x + 1)$$ 4. **Write the fraction and simplify:** $$\frac{f(x)}{f(x+3)} = \frac{(x - 2)(x + 1)}{(x + 4)(x + 1)}$$ Cancel the common factor $(x + 1)$ (not zero to avoid division by zero): $$\frac{(x - 2)\cancel{(x + 1)}}{(x + 4)\cancel{(x + 1)}}$$ So, $$\frac{f(x)}{f(x+3)} = \frac{x - 2}{x + 4}$$ 5. **Condition for simplification:** $f(x+3) \neq 0$ means $(x + 4)(x + 1) \neq 0$, so $x \neq -4$ and $x \neq -1$. 6. **Solve inequality:** $$\frac{f(x)}{f(x+3)} < 2 \implies \frac{x - 2}{x + 4} < 2$$ Rewrite: $$\frac{x - 2}{x + 4} - 2 < 0$$ $$\frac{x - 2 - 2(x + 4)}{x + 4} < 0$$ Simplify numerator: $$x - 2 - 2x - 8 = -x - 10$$ So inequality is: $$\frac{-x - 10}{x + 4} < 0$$ Or equivalently: $$\frac{-(x + 10)}{x + 4} < 0$$ 7. **Analyze sign:** The fraction is negative when numerator and denominator have opposite signs. - Numerator $-(x + 10)$ is positive when $x + 10 < 0 \Rightarrow x < -10$. - Numerator is negative when $x > -10$. - Denominator $x + 4$ is positive when $x > -4$. - Denominator is negative when $x < -4$. 8. **Check intervals:** - For $x < -10$: numerator positive, denominator negative (since $x < -4$), fraction negative $\Rightarrow$ inequality holds. - For $-10 < x < -4$: numerator negative, denominator negative, fraction positive $\Rightarrow$ inequality does not hold. - For $x > -4$: numerator negative, denominator positive, fraction negative $\Rightarrow$ inequality holds. 9. **Exclude points where denominator zero:** $x \neq -4$ (given), also $x \neq -1$ from simplification step. 10. **Final solution:** $$x < -10 \quad \text{or} \quad x > -4, \quad x \neq -1$$ **Answer:** (i) $\frac{f(x)}{f(x+3)} = \frac{x - 2}{x + 4}$ for $x \neq -4, -1$. (ii) $\frac{f(x)}{f(x+3)} < 2$ for $x < -10$ or $x > -4$ with $x \neq -1$.