1. **State the problem:** Find the domain and range of the piecewise function: $$q(x) = \begin{cases} \sqrt{x+3} & \text{if } -3 \leq x < 1 \\ |x-2|+1 & \text{if } 1 \leq x \leq 5 \\ 2^{x-5} & \text{if } x > 5 \end{cases}$$ 2. **Find the domain:** The domain is all $x$ values for which the function is defined. - For $\sqrt{x+3}$, the radicand must be $\geq 0$, so $x+3 \geq 0 \Rightarrow x \geq -3$. Given $-3 \leq x < 1$, this part is valid. - For $|x-2|+1$, absolute value is defined for all real numbers, so domain is $1 \leq x \leq 5$. - For $2^{x-5}$, exponential functions are defined for all real $x$, so domain is $x > 5$. Combining all intervals, the domain is: $$[-3,1) \cup [1,5] \cup (5, \infty) = [-3, \infty)$$ 3. **Find the range:** Analyze each piece separately. - For $\sqrt{x+3}$ on $[-3,1)$: - Minimum at $x=-3$: $\sqrt{-3+3} = \sqrt{0} = 0$ - Maximum as $x \to 1^-$: $\sqrt{1+3} = \sqrt{4} = 2$ - Range is $[0,2)$ because $x=1$ is not included. - For $|x-2|+1$ on $[1,5]$: - $|x-2|$ is minimum at $x=2$ with value 0, so minimum of this piece is $0+1=1$. - At $x=1$, $|1-2|+1 = 1+1=2$. - At $x=5$, $|5-2|+1 = 3+1=4$. - The function forms a "V" shape with minimum 1 at $x=2$ and maximum 4 at $x=5$. - Range is $[1,4]$. - For $2^{x-5}$ on $(5, \infty)$: - At $x=5$, $2^{0} = 1$ (not included since $x>5$). - As $x \to \infty$, $2^{x-5} \to \infty$. - Range is $(1, \infty)$. 4. **Combine ranges:** - From first piece: $[0,2)$ - Second piece: $[1,4]$ - Third piece: $(1, \infty)$ The union is: $$[0,2) \cup [1,4] \cup (1, \infty) = [0, \infty)$$ 5. **Final answers:** - Domain: $[-3, \infty)$ - Range: $[0, \infty)$