1. **Problem:** Decide whether the given relations are functions or not and justify your answer. **a)** $y^2 = x - 3$ - To check if this relation is a function, solve for $y$: $$y = \pm \sqrt{x - 3}$$ - For each $x \geq 3$, there are two possible $y$ values (positive and negative square roots). - Since a function must assign exactly one $y$ for each $x$, this relation is **not a function**. **b)** $y = -x^4 + 7$ - This is an explicit function of $x$. - For each $x$, there is exactly one $y$. - Therefore, this relation **is a function**. 2. **Problem:** Determine algebraically whether the following functions are even, odd, or neither. Recall: - A function $f$ is **even** if $f(-x) = f(x)$ for all $x$. - A function $f$ is **odd** if $f(-x) = -f(x)$ for all $x$. **a)** $y = x - 3$ - Compute $f(-x) = -x - 3$ - Compare with $f(x) = x - 3$ - Since $f(-x) \neq f(x)$ and $f(-x) \neq -f(x)$, the function is **neither even nor odd**. **b)** $y = -x^4 + 7$ - Compute $f(-x) = -(-x)^4 + 7 = -x^4 + 7 = f(x)$ - Since $f(-x) = f(x)$, the function is **even**. **c)** $f(x) = -x^3 - x$ - Compute $f(-x) = -(-x)^3 - (-x) = -(-x^3) + x = x^3 + x$ - Compare with $-f(x) = -(-x^3 - x) = x^3 + x$ - Since $f(-x) = -f(x)$, the function is **odd**. 3. **Problem:** Solve the inequalities. **a)** $-8 \leq \frac{x}{6} \leq 1$ - Multiply all parts by 6 (positive, so inequality signs stay the same): $$-8 \times 6 \leq x \leq 1 \times 6$$ $$-48 \leq x \leq 6$$ **b)** $-3 \leq 1 - \frac{2x}{5} \leq 3$ - Subtract 1 from all parts: $$-3 - 1 \leq -\frac{2x}{5} \leq 3 - 1$$ $$-4 \leq -\frac{2x}{5} \leq 2$$ - Multiply all parts by $-\frac{5}{2}$ (negative, so reverse inequalities): $$-4 \times -\frac{5}{2} \geq x \geq 2 \times -\frac{5}{2}$$ $$10 \geq x \geq -5$$ - Rewrite as: $$-5 \leq x \leq 10$$ 4. **Problem:** Determine whether **c)** $|3x + 4| = |2x + 1|$ - Square both sides to remove absolute values: $$(3x + 4)^2 = (2x + 1)^2$$ - Expand: $$9x^2 + 24x + 16 = 4x^2 + 4x + 1$$ - Bring all terms to one side: $$9x^2 + 24x + 16 - 4x^2 - 4x - 1 = 0$$ $$5x^2 + 20x + 15 = 0$$ - Divide by 5: $$x^2 + 4x + 3 = 0$$ - Factor: $$(x + 3)(x + 1) = 0$$ - Solutions: $$x = -3, -1$$ **d)** $3(4 - y) \geq 9$ - Distribute: $$12 - 3y \geq 9$$ - Subtract 12: $$-3y \geq -3$$ - Divide by -3 (reverse inequality): $$y \leq 1$$ **Final answers:** - Exercise 1: a) Not a function, b) Function - Exercise 2: a) Neither, b) Even, c) Odd - Exercise 3: a) $-48 \leq x \leq 6$, b) $-5 \leq x \leq 10$ - Exercise 4: c) $x = -3, -1$, d) $y \leq 1$