1. Problem: Calculate $4f(-1) + 3g(-2)$ given $f(x) = 3x - 7$ and $g(x) = 4x + 2$. 2. First, find $f(-1)$: $$f(-1) = 3(-1) - 7 = -3 - 7 = -10.$$ 3. Next, find $g(-2)$: $$g(-2) = 4(-2) + 2 = -8 + 2 = -6.$$ 4. Now substitute into the expression: $$4f(-1) + 3g(-2) = 4(-10) + 3(-6) = -40 - 18 = -58.$$ --- 5. Problem: For $f(x) = \frac{4x + 1}{3x - 2}$, $x \neq \frac{2}{3}$, find: i. $f^{-1}(2)$ ii. $f(f(\frac{1}{2}))$ iii. $f(f^{-1}(4))$ 6. Find the inverse function $f^{-1}(x)$: Start with $y = \frac{4x + 1}{3x - 2}$. Solve for $x$: $$y(3x - 2) = 4x + 1 \implies 3yx - 2y = 4x + 1$$ $$3yx - 4x = 2y + 1 \implies x(3y - 4) = 2y + 1$$ $$x = \frac{2y + 1}{3y - 4}.$$ Interchange $x$ and $y$ to write inverse: $$f^{-1}(x) = \frac{2x + 1}{3x - 4}, \quad x \neq \frac{4}{3}.$$ 7. i. Evaluate $f^{-1}(2)$: $$f^{-1}(2) = \frac{2(2) + 1}{3(2) - 4} = \frac{4 + 1}{6 - 4} = \frac{5}{2}.$$ 8. ii. Evaluate $f(\frac{1}{2})$: $$f(\frac{1}{2}) = \frac{4\cdot \frac{1}{2} + 1}{3 \cdot \frac{1}{2} - 2} = \frac{2 + 1}{1.5 - 2} = \frac{3}{-0.5} = -6.$$ Then, find $f(f(\frac{1}{2})) = f(-6)$: $$f(-6) = \frac{4(-6) + 1}{3(-6) - 2} = \frac{-24 + 1}{-18 - 2} = \frac{-23}{-20} = \frac{23}{20}.$$ 9. iii. Simplify $f(f^{-1}(4))$: Since $f$ and $f^{-1}$ are inverses, $$f(f^{-1}(4)) = 4.$$ --- 10. Problem: Given $f(x) = \frac{4x}{x + 1}$, $x \neq k$ and $g(x) = x + 4$, find: i. Value of $k$. ii. Expression for $fg(x)$. iii. Inverse function $f^{-1}(x)$. iv. Value of $a$ such that $f^{-1}(a) = g(-1)$. 11. i. The domain restriction $x \neq k$ arises because denominator $x + 1$ cannot be zero: $$x + 1 \neq 0 \Rightarrow x \neq -1.$$ So, $k = -1$. 12. ii. Find $fg(x) = f(g(x))$: $$g(x) = x + 4.$$ Substitute: $$fg(x) = f(x+4) = \frac{4(x+4)}{(x+4) + 1} = \frac{4x + 16}{x + 5}.$$ 13. iii. Find inverse $f^{-1}(x)$: Start with $$y = \frac{4x}{x+1}.$$ Solve for $x$: $$y(x+1) = 4x \, \Rightarrow \, yx + y = 4x$$ $$yx - 4x = -y \, \Rightarrow \, x(y - 4) = -y$$ $$x = \frac{-y}{y - 4} = \frac{y}{4 - y}$$ Interchange $x$ and $y$ gives: $$f^{-1}(x) = \frac{x}{4 - x}, \quad x \neq 4.$$ 14. iv. Find $a$ such that $$f^{-1}(a) = g(-1).$$ Calculate $g(-1)$: $$g(-1) = -1 + 4 = 3.$$ Set $$f^{-1}(a) = 3 \Rightarrow \frac{a}{4 - a} = 3.$$ Solve for $a$: $$a = 3(4 - a) = 12 - 3a$$ $$a + 3a = 12 \Rightarrow 4a = 12 \Rightarrow a = 3.$$ --- 15. Problem: Given $$f(x) = \frac{2 - x}{x + 3}$$ Find values of $x$ for which $$f(x) > 1.$$ 16. Solve $$\frac{2 - x}{x + 3} > 1.$$ Since denominator cannot be zero, $$x \neq -3.$$ Rewrite inequality: $$\frac{2 - x}{x + 3} - 1 > 0 \Rightarrow \frac{2 - x - (x + 3)}{x + 3} > 0.$$ Simplify numerator: $$2 - x - x - 3 = -2x - 1.$$ So, inequality is: $$\frac{-2x - 1}{x + 3} > 0.$$ 17. Find critical points from numerator and denominator: Numerator zero: $$-2x - 1 = 0 \Rightarrow x = -\frac{1}{2}.$$ Denominator zero: $$x = -3.$$ 18. Test intervals divided by points $-3$ and $-\frac{1}{2}$: - For $x < -3$: Numerator negative (e.g., $x = -4$: $-2(-4)-1=8-1=7 > 0$ actually positive), denominator negative. So ratio positive over negative = negative $< 0$. So NO. - For $-3 < x < -\frac{1}{2}$: Numerator at $x=-2$: $-2(-2)-1=4-1=3>0$. Denominator positive. Ratio positive/positive = positive $> 0$. YES. - For $x > -\frac{1}{2}$: Numerator at $x=0$: $-2(0)-1=-1<0$. Denominator positive. Ratio negative/positive $<0$. NO. 19. Domain excludes $x = -3$, so solution is: $$-3 < x < -\frac{1}{2}.$$ --- 20. Problem: Solve $$3x^2 - 19|x| + 20 = 0.$$ 21. Substitute $t = |x| \geq 0$, then solve quadratic in $t$: $$3t^2 - 19t + 20 = 0.$$ 22. Use quadratic formula for $t$: $$t = \frac{19 \pm \sqrt{(-19)^2 - 4 \cdot 3 \cdot 20}}{2 \cdot 3} = \frac{19 \pm \sqrt{361 - 240}}{6} = \frac{19 \pm \sqrt{121}}{6} = \frac{19 \pm 11}{6}.$$ 23. Calculate roots: $$t_1 = \frac{19 + 11}{6} = \frac{30}{6} = 5,$$ $$t_2 = \frac{19 - 11}{6} = \frac{8}{6} = \frac{4}{3}.$$ 24. Recall $t = |x|$, so $$|x| = 5 \Rightarrow x = \pm 5,$$ $$|x| = \frac{4}{3} \Rightarrow x = \pm \frac{4}{3}.$$ 25. Final solution set: $$x = \pm 5, \pm \frac{4}{3}.$$