1. Problem: Given functions and piecewise definitions, find expressions and values for g(x), f(g(11)), and solve functional equations. 2. For the first problem, f(x) = 3x + 1 and g(x) is defined piecewise as: $$g(x) = \begin{cases} 3x + 10, & x < 3 \\ 3x - 8, & x \geq 3 \end{cases}$$ We want to find which expression g(x) equals among options involving absolute values. 3. Consider the function f(x) = 3x + 1. To express g(x) in terms of absolute values, check the behavior at x=3: - For x < 3, g(x) = 3x + 10 - For x \geq 3, g(x) = 3x - 8 4. Test option D) $|x - 3|$ scaled by 3: $$3|x - 3| = \begin{cases} 3(3 - x) = 9 - 3x, & x < 3 \\ 3(x - 3) = 3x - 9, & x \geq 3 \end{cases}$$ This does not match g(x). 5. Test option A) $3|x - 1|$: $$3|x - 1| = \begin{cases} 3(1 - x) = 3 - 3x, & x < 1 \\ 3(x - 1) = 3x - 3, & x \geq 1 \end{cases}$$ Does not match g(x). 6. Test option B) $|x| - 3$: $$|x| - 3$$ is continuous but does not match the piecewise linear form of g(x). 7. Test option C) $3|x| - 1$: $$3|x| - 1$$ is symmetric about zero, not matching g(x). 8. None of the options exactly match g(x) as given, so g(x) is as defined piecewise. 9. Next, given f(x) = 2x - 3, find g(x). The problem is incomplete for g(x) here; assuming g(x) is the same piecewise function or needs definition. 10. Calculate f(g(11)): Since 11 \geq 3, use $g(11) = 3(11) - 8 = 33 - 8 = 25$. Then, $$f(g(11)) = f(25) = 2(25) - 3 = 50 - 3 = 47$$ 11. For the rational expressions: Given $$\frac{2x + 5}{3} = 2x^2 + 5x - 2$$ and $$\frac{2x + 5}{3} = 4x^2 - 6x + 3$$ These are inconsistent equalities; likely a misunderstanding or separate equations. 12. For function $$f(x) = \frac{x}{x} + x^2 - x$$ Simplify: $$\frac{x}{x} = 1 \quad (x \neq 0)$$ So, $$f(x) = 1 + x^2 - x$$ 13. For $$f(x) = 4^{3x^{-2}}$$ and $$g(x) = \log_2(x + 1)$$ Calculate $$f(g(3))$$ First, $$g(3) = \log_2(3 + 1) = \log_2(4) = 2$$ Then, $$f(g(3)) = 4^{3 \cdot 2^{-2}} = 4^{3 \cdot \frac{1}{4}} = 4^{\frac{3}{4}}$$ Since $$4 = 2^2,$$ we have $$4^{\frac{3}{4}} = (2^2)^{\frac{3}{4}} = 2^{2 \cdot \frac{3}{4}} = 2^{\frac{3}{2}} = 2^{1.5} = 2 \sqrt{2}$$ 14. For the recurrence relation: $$f(x + 1) = 3f(x) - f(x - 1) + 2x$$ with initial values $$f(1) = 6, \quad f(2) = 5$$ Find $f(5)$. Calculate stepwise: - $f(3) = 3f(2) - f(1) + 2 \cdot 2 = 3 \cdot 5 - 6 + 4 = 15 - 6 + 4 = 13$ - $f(4) = 3f(3) - f(2) + 2 \cdot 3 = 3 \cdot 13 - 5 + 6 = 39 - 5 + 6 = 40$ - $f(5) = 3f(4) - f(3) + 2 \cdot 4 = 3 \cdot 40 - 13 + 8 = 120 - 13 + 8 = 115$ Answer: $f(5) = 115$ 15. Summary answers: - $g(x)$ is the given piecewise function. - $f(g(11)) = 47$ - Simplified $f(x) = 1 + x^2 - x$ - $f(g(3)) = 2 \sqrt{2}$ - $f(5) = 115$