1. **Problem statement:** Given real numbers $x$ and $y$ such that $2 \leq x \leq 5$ and $-3 \leq y \leq 4$, find the bounds (range) for the expressions $x+y$, $x-y$, $xy$, and $y^2$. 2. **Formula and rules:** To find the bounds of sums, differences, products, and powers, use the given intervals and consider the minimum and maximum values by evaluating at the interval endpoints. 3. **Step-by-step solution:** - For $x+y$: - Minimum: $2 + (-3) = -1$ - Maximum: $5 + 4 = 9$ - So, $-1 \leq x+y \leq 9$ - For $x-y$: - Minimum: $2 - 4 = -2$ - Maximum: $5 - (-3) = 8$ - So, $-2 \leq x-y \leq 8$ - For $xy$: - Since $x$ is positive and $y$ can be negative or positive, check all combinations: - $2 \times (-3) = -6$ - $2 \times 4 = 8$ - $5 \times (-3) = -15$ - $5 \times 4 = 20$ - Minimum is $-15$, maximum is $20$ - So, $-15 \leq xy \leq 20$ - For $y^2$: - Since $y$ ranges from $-3$ to $4$, square values range from $0$ to $16$ (because $(-3)^2=9$ and $4^2=16$) - Minimum is $0$ (at $y=0$), maximum is $16$ - So, $0 \leq y^2 \leq 16$ **Final answer:** $$ \begin{cases} -1 \leq x+y \leq 9 \\ -2 \leq x-y \leq 8 \\ -15 \leq xy \leq 20 \\ 0 \leq y^2 \leq 16 \end{cases} $$