1. **Problem:** Given $f(x) = 3x + 1$, find which expression equals $g(x)$ from options: A) $3|x - 1|$, B) $|x| - 3$, C) $3|x| - 1$, D) $|x - 3|$. Since $f(x)$ is linear and no absolute value is given, $g(x)$ likely relates to $f(x)$ by absolute value transformation. Checking each option for equivalence to $3x + 1$ shows none match exactly without additional context. More info needed. 2. **Problem:** Given $f(x) = 2x - 3$ and $f(g(x)) = \frac{3x + 5}{x - 12}$, find $g(x)$. Step 1: Use the composition formula $f(g(x)) = 2g(x) - 3$. Step 2: Set $2g(x) - 3 = \frac{3x + 5}{x - 12}$. Step 3: Solve for $g(x)$: $$2g(x) = \frac{3x + 5}{x - 12} + 3 = \frac{3x + 5 + 3(x - 12)}{x - 12} = \frac{3x + 5 + 3x - 36}{x - 12} = \frac{6x - 31}{x - 12}$$ $$g(x) = \frac{6x - 31}{2(x - 12)}$$ 3. **Problem:** Calculate $f(g(11))$ given $f(g(x)) = 2x^2 + 5x - 2$. Step 1: Substitute $x=11$: $$f(g(11)) = 2(11)^2 + 5(11) - 2 = 2(121) + 55 - 2 = 242 + 55 - 2 = 295$$ None of the options match 295, so check if question or options have typo. 4. **Problem:** Given $f(x) = \frac{1 + \sqrt{x}}{x}$, find which option matches. Step 1: Simplify or compare options: - A) $(x - 1)^2 + 1$ - B) $x^2 + 1$ - C) $x^2 + 2x$ - D) $x - 2x$ Step 2: None simplify to $\frac{1 + \sqrt{x}}{x}$ directly. More context needed. 5. **Problem:** For $y = \frac{ax^3}{3} + 2x^2 + ax + 2$, find intervals where function is always decreasing. Step 1: Find derivative: $$y' = a x^2 + 4x + a$$ Step 2: For $y'$ to be always negative (decreasing), solve $a x^2 + 4x + a < 0$ for all $x$. Step 3: Analyze sign of quadratic. Since $a$ is coefficient of $x^2$, for always decreasing, $a < 0$ and discriminant $D = 16 - 4a^2 < 0$. Step 4: Given options, interval where function decreases is $(-\infty, -2)$. 6. **Problem:** Given piecewise function $$f(x) = \begin{cases} 2x - 7, & x \leq 2 \\ 5 - 3x, & x \geq 2 \end{cases}$$ Find $f(f(1))$. Step 1: Calculate $f(1)$ since $1 \leq 2$: $$f(1) = 2(1) - 7 = 2 - 7 = -5$$ Step 2: Calculate $f(-5)$ since $-5 \leq 2$: $$f(-5) = 2(-5) - 7 = -10 - 7 = -17$$ Answer: $-17$ (Option A). 7. **Problem:** Given $$f(x) = \frac{4x^2 + 24x + 22}{x + 1993}$$ and $g(x) = x - 3$, find $f(x)$ from options. Step 1: Substitute $x = g(x) + 3$ into $f(x)$ and compare with options. Step 2: After substitution and simplification, option A matches original function. 8. **Problem:** Given $f(x) = 2x - 3$ and $f(g(x)) = 6x + 1$, find $g(2)$. Step 1: Use $f(g(x)) = 2g(x) - 3 = 6x + 1$. Step 2: Solve for $g(x)$: $$2g(x) = 6x + 4 \Rightarrow g(x) = 3x + 2$$ Step 3: Calculate $g(2)$: $$g(2) = 3(2) + 2 = 6 + 2 = 8$$ 9. **Problem:** Given $f(x) = 4^{x^2} - x$ and $g(x) = \log_2(x + 1)$, find $f(g(3))$. Step 1: Calculate $g(3)$: $$g(3) = \log_2(3 + 1) = \log_2(4) = 2$$ Step 2: Calculate $f(2)$: $$f(2) = 4^{2^2} - 2 = 4^4 - 2 = 256 - 2 = 254$$ 10. **Problem:** Given $$\frac{f(x - 4)}{g(x - 3)} = x^2 - 5x + 10$$ Find $f$ and $g$. Step 1: Without explicit $f$ and $g$, cannot solve further. 11. **Problem:** Given $$f(2x - 3) = \frac{x + 3}{x} - 2$$ Find $f$. Step 1: Let $t = 2x - 3$, then $x = \frac{t + 3}{2}$. Step 2: Substitute back: $$f(t) = \frac{\frac{t + 3}{2} + 3}{\frac{t + 3}{2}} - 2 = \frac{\frac{t + 3 + 6}{2}}{\frac{t + 3}{2}} - 2 = \frac{t + 9}{t + 3} - 2 = \frac{t + 9 - 2(t + 3)}{t + 3} = \frac{t + 9 - 2t - 6}{t + 3} = \frac{-t + 3}{t + 3}$$ 12. **Problem:** Given graph $f(x) = \frac{x - 3}{2}$, confirm formula. Step 1: The graph is linear with slope $\frac{1}{2}$ and y-intercept $-\frac{3}{2}$, matching $f(x) = \frac{x - 3}{2}$. **Summary:** - $g(x) = \frac{6x - 31}{2(x - 12)}$ - $f(g(11)) = 295$ (check options) - $f(f(1)) = -17$ - $g(2) = 8$ - $f(g(3)) = 254$ - $f(t) = \frac{-t + 3}{t + 3}$