1. **Find the domain of** $f(x) = x^2 + \frac{x}{5} + \sqrt[3]{x - 11}$. - The function includes a cube root $\sqrt[3]{x - 11}$ which is defined for all real numbers. - The polynomial terms $x^2$ and $\frac{x}{5}$ are defined for all real numbers. - Therefore, the only restriction is from the cube root's radicand $x - 11$ if it were an even root, but since it is a cube root, no restriction. - However, the problem states domain $(-11, \infty)$, which seems to be a mistake; cube roots are defined for all real $x$. **Correct domain:** $(-\infty, \infty)$. 2. **Find the domain of** $y = \frac{5x^2 + x}{3x - 6}$. - The denominator cannot be zero. - Set denominator equal to zero: $3x - 6 = 0 \Rightarrow x = 2$. - So domain is all real numbers except $x = 2$. **Domain:** $(-\infty, 2) \cup (2, \infty)$. 3. **Find the domain of** $\sqrt{-4x + 8}$. - The radicand must be $\geq 0$. - Solve $-4x + 8 \geq 0 \Rightarrow -4x \geq -8 \Rightarrow x \leq 2$. **Domain:** $(-\infty, 2]$. 4. **Find intercepts of** $y = 4x^2 - 2x$. - $y$-intercept: set $x=0$, $y=0$ so $(0,0)$. - $x$-intercepts: set $y=0$, solve $4x^2 - 2x = 0$. - Factor: $2x(2x - 1) = 0$. - Solutions: $x=0$ and $x=\frac{1}{2}$. - So $x$-intercepts are $(0,0)$ and $(\frac{1}{2},0)$. 5. **Find intercepts of** $y = \sqrt{2x + 10}$. - $y$-intercept: set $x=0$, $y=\sqrt{10}$, so $(0, \sqrt{10})$. - $x$-intercept: set $y=0$, solve $\sqrt{2x + 10} = 0 \Rightarrow 2x + 10 = 0 \Rightarrow x = -5$. - So $x$-intercept is $(-5,0)$. 6. **Find** $(f + g)(x)$ where $f(x) = -3x + 4$, $g(x) = 2x - 1$. - $(f + g)(x) = f(x) + g(x) = (-3x + 4) + (2x - 1) = -3x + 4 + 2x - 1 = (-3x + 2x) + (4 - 1) = -x + 3$. 7. **Find** $(g - f)(x)$. - $(g - f)(x) = g(x) - f(x) = (2x - 1) - (-3x + 4) = 2x - 1 + 3x - 4 = (2x + 3x) + (-1 - 4) = 5x - 5$. 8. **Find** $(fg)(x)$ (product). - $(fg)(x) = f(x) \cdot g(x) = (-3x + 4)(2x - 1)$. - Multiply: $-3x \cdot 2x = -6x^2$, $-3x \cdot (-1) = +3x$, $4 \cdot 2x = 8x$, $4 \cdot (-1) = -4$. - Sum: $-6x^2 + 3x + 8x - 4 = -6x^2 + 11x - 4$. 9. **Find** $\left(\frac{g}{f}\right)(x)$. - $\frac{g(x)}{f(x)} = \frac{2x - 1}{-3x + 4}$. 10. **Find** $(f \circ g)(x) = f(g(x))$. - $g(x) = 2x - 1$. - Substitute into $f$: $f(g(x)) = -3(2x - 1) + 4 = -6x + 3 + 4 = -6x + 7$. 11. **Find** $(g \circ f)(x) = g(f(x))$. - $f(x) = -3x + 4$. - Substitute into $g$: $g(f(x)) = 2(-3x + 4) - 1 = -6x + 8 - 1 = -6x + 7$. 12. **Find** $(f \circ f)(-3)$. - First find $f(-3) = -3(-3) + 4 = 9 + 4 = 13$. - Then find $f(13) = -3(13) + 4 = -39 + 4 = -35$. 13. **Determine if** $f(x) = 2x - 9$ and $g(x) = \frac{x + 9}{2}$ are inverses. - Compute $f(g(x)) = 2\left(\frac{x + 9}{2}\right) - 9 = (x + 9) - 9 = x$. - Compute $g(f(x)) = \frac{2x - 9 + 9}{2} = \frac{2x}{2} = x$. - Since both compositions equal $x$, $f$ and $g$ are inverses. **Final answers:** - Domains: 1. $(-\infty, \infty)$ 2. $(-\infty, 2) \cup (2, \infty)$ 3. $(-\infty, 2]$ - Intercepts: 4. $y$-int: $(0,0)$; $x$-int: $(0,0), (\frac{1}{2}, 0)$ 5. $y$-int: $(0, \sqrt{10})$; $x$-int: $(-5,0)$ - Function operations: 6. $(f + g)(x) = -x + 3$ 7. $(g - f)(x) = 5x - 5$ 8. $(fg)(x) = -6x^2 + 11x - 4$ 9. $\left(\frac{g}{f}\right)(x) = \frac{2x - 1}{-3x + 4}$ 10. $(f \circ g)(x) = -6x + 7$ 11. $(g \circ f)(x) = -6x + 7$ 12. $(f \circ f)(-3) = -35$ - Inverse check: 13. $f$ and $g$ are inverses.