1. **State the problem:** Find the symmetry of the function $f(x) = \frac{x^2 - 4}{5 + x}$. 2. **Recall the rules for symmetry:** - A function is even if $f(-x) = f(x)$ for all $x$ in the domain. - A function is odd if $f(-x) = -f(x)$ for all $x$ in the domain. - Otherwise, the function has no symmetry. 3. **Calculate $f(-x)$:** $$f(-x) = \frac{(-x)^2 - 4}{5 + (-x)} = \frac{x^2 - 4}{5 - x}$$ 4. **Compare $f(-x)$ with $f(x)$:** $$f(x) = \frac{x^2 - 4}{5 + x}$$ $$f(-x) = \frac{x^2 - 4}{5 - x}$$ Since $f(-x) \neq f(x)$, the function is not even. 5. **Check if $f(-x) = -f(x)$:** Calculate $-f(x)$: $$-f(x) = -\frac{x^2 - 4}{5 + x} = \frac{-(x^2 - 4)}{5 + x} = \frac{-x^2 + 4}{5 + x}$$ Since $f(-x) = \frac{x^2 - 4}{5 - x}$ and $-f(x) = \frac{-x^2 + 4}{5 + x}$, they are not equal. 6. **Conclusion:** The function $f(x) = \frac{x^2 - 4}{5 + x}$ is neither even nor odd; it has no symmetry. --- 1. **State the problem:** Find the symmetry of the function $f(x) = \frac{4}{x^2 + x - 4}$. 2. **Recall the rules for symmetry:** (same as above) 3. **Calculate $f(-x)$:** $$f(-x) = \frac{4}{(-x)^2 + (-x) - 4} = \frac{4}{x^2 - x - 4}$$ 4. **Compare $f(-x)$ with $f(x)$:** $$f(x) = \frac{4}{x^2 + x - 4}$$ $$f(-x) = \frac{4}{x^2 - x - 4}$$ Since $f(-x) \neq f(x)$, the function is not even. 5. **Check if $f(-x) = -f(x)$:** Calculate $-f(x)$: $$-f(x) = -\frac{4}{x^2 + x - 4} = \frac{-4}{x^2 + x - 4}$$ Since $f(-x) = \frac{4}{x^2 - x - 4}$ and $-f(x) = \frac{-4}{x^2 + x - 4}$, they are not equal. 6. **Conclusion:** The function $f(x) = \frac{4}{x^2 + x - 4}$ is neither even nor odd; it has no symmetry. **Final answers:** - (a) No symmetry - (b) No symmetry