1. **Problem:** Find the domain and range of $f(x) = x^2$. 2. **Domain:** Since $f(x) = x^2$ has no denominators or square roots, all real numbers are allowed. Thus, domain is $\mathbb{R}$. 3. **Range:** Squaring any real number results in a nonnegative value. The minimum value is 0 at $x=0$, and values increase without bound as $|x|$ grows. Hence, range is $[0, \infty)$. 4. **Final Answer:** Domain $= \mathbb{R}$, Range $= [0, \infty)$. --- 1. **Problem:** Find the domain of $f(x) = \frac{1}{x - 2}$. 2. **Domain restriction:** The denominator cannot be zero, so $x - 2 \neq 0$. 3. **Solve:** $x \neq 2$. 4. **Final Answer:** Domain $= \mathbb{R} \setminus \{2\}$. --- 1. **Problem:** Find the domain and range of $f(x) = \sqrt{x + 3}$. 2. **Domain:** The expression inside the square root must be nonnegative: $x + 3 \geq 0 \Rightarrow x \geq -3$. 3. **Range:** Square root outputs are always $\geq 0$. Minimum value is 0 at $x = -3$, and values increase without bound as $x$ increases. 4. **Final Answer:** Domain $= [-3, \infty)$, Range $= [0, \infty)$. --- 1. **Problem:** Given $f(x) = x + 1$ and $g(x) = x^2$, find $(f \circ g)(2)$ and $(g \circ f)(2)$. 2. **Compute $(f \circ g)(2)$:** - First find $g(2) = 2^2 = 4$. - Then $f(g(2)) = f(4) = 4 + 1 = 5$. 3. **Compute $(g \circ f)(2)$:** - First find $f(2) = 2 + 1 = 3$. - Then $g(f(2)) = g(3) = 3^2 = 9$. 4. **Final Answer:** $(f \circ g)(2) = 5$, $(g \circ f)(2) = 9$. --- 1. **Problem:** Find the inverse of $f(x) = 3x + 4$ and verify by composition. 2. **Find inverse:** Let $y = 3x + 4$. Swap $x$ and $y$: $x = 3y + 4$. 3. **Solve for $y$:** $3y = x - 4 \Rightarrow y = \frac{x - 4}{3}$. 4. **Inverse function:** $f^{-1}(x) = \frac{x - 4}{3}$. 5. **Verification:** - Compute $f(f^{-1}(x)) = 3\left(\frac{x - 4}{3}\right) + 4 = x - 4 + 4 = x$. 6. **Final Answer:** $f^{-1}(x) = \frac{x - 4}{3}$. --- 1. **Problem:** Find the inverse of $f(x) = \frac{x - 2}{5}$ and verify by composition. 2. **Find inverse:** Let $y = \frac{x - 2}{5}$. Swap $x$ and $y$: $x = \frac{y - 2}{5}$. 3. **Solve for $y$:** Multiply both sides by 5: $5x = y - 2 \Rightarrow y = 5x + 2$. 4. **Inverse function:** $f^{-1}(x) = 5x + 2$. 5. **Verification:** - Compute $f(f^{-1}(x)) = \frac{(5x + 2) - 2}{5} = \frac{5x}{5} = x$. 6. **Final Answer:** $f^{-1}(x) = 5x + 2$.