Question:

Step 1: Analyze $f(x) = \sqrt{1 - x^2}$ For the domain, the expression under the square root must be non-negative: $$1 - x^2 \ge 0 \implies x^2 \le 1 \implies -1 \le x \le 1$$ The domain of $f(x)$ is $[-1, 1]$. For the range, let $y = \sqrt{1 - x^2}$. Since $x \in [-1, 1]$, we have $x^2 \in [0, 1]$. Therefore, $1 - x^2 \in [0, 1]$. Taking the square root, $y = \sqrt{1 - x^2} \in [0, 1]$. The range of $f(x)$ is $[0, 1]$. The domain $[-1, 1]$ is not equal to the range $[0, 1]$. Step 2: Analyze $g(x) = \frac{1}{x}$ For the domain, the denominator cannot be zero: $$x \neq 0$$ The domain of $g(x)$ is $\mathbb{R} \setminus \{0\}$, which can be written as $(-\infty, 0) \cup (0, \infty)$. For the range, let $y = \frac{1}{x}$. For any non-zero real number $y$, we can find an $x$ such that $y = \frac{1}{x}$, specifically $x = \frac{1}{y}$. The range of $g(x)$ is $\mathbb{R} \setminus \{0\}$, which can be written as $(-\infty, 0) \cup (0, \infty)$. The domain $\mathbb{R} \setminus \{0\}$ is equal to the range $\mathbb{R} \setminus \{0\}$. Step 3: Analyze $h(x) = \sqrt{x}$ For the domain, the expression under the square root must be non-negative: $$x \ge 0$$ The domain of $h(x)$ is $[0, \infty)$. For the range, the square root function $\sqrt{x}$ produces all non-negative real values when $x \ge 0$. The range of $h(x)$ is $[0, \infty)$. The domain $[0, \infty)$ is equal to the range $[0, \infty)$. Step 4: Analyze $l(x) = \sqrt{4 - x}$ For the domain, the expression under the square root must be non-negative: $$4 - x \ge 0 \implies x \le 4$$ The domain of $l(x)$ is $(-\infty, 4]$. For the range, let $y = \sqrt{4 - x}$. As $x$ takes values in $(-\infty, 4]$, the expression $4 - x$ takes values in $[0, \infty)$. Therefore, $y = \sqrt{4 - x}$ takes values in $[0, \infty)$. The range of $l(x)$ is $[0, \infty)$. The domain $(-\infty, 4]$ is not equal to the range $[0, \infty)$. Step 5: Conclusion The functions that have the same domain and range are $g(x) = \frac{1}{x}$ and $h(x) = \sqrt{x}$.