1) Given the function $f(x) = 2x - 5$: 1. a) Find $f'(x)$, the derivative of $f(x)$. Since $f(x) = 2x - 5$, the derivative is the rate of change of $f$ with respect to $x$. $$f'(x) = \frac{d}{dx}(2x - 5) = 2$$ 1. b) Find $f^{-1}(2)$, the inverse function's value at 2. To find $f^{-1}(2)$, solve $f(x) = 2$ for $x$: $$2x - 5 = 2$$ $$2x = 7$$ $$x = \frac{7}{2} = 3.5$$ 1. c) Find $x$ if $f(f(f(x)))=x$. Calculate the triple composition: $$f(x) = 2x - 5$$ $$f(f(x)) = f(2x - 5) = 2(2x - 5) - 5 = 4x - 10 - 5 = 4x - 15$$ $$f(f(f(x))) = f(4x - 15) = 2(4x - 15) - 5 = 8x - 30 - 5 = 8x - 35$$ Set equal to $x$: $$8x - 35 = x$$ $$8x - x = 35$$ $$7x = 35$$ $$x = 5$$ 2) Evaluate $\left(\frac{8}{27}\right)^{1/3}$. Recall: $a^{1/3}$ is the cube root of $a$. $$\left(\frac{8}{27}\right)^{1/3} = \frac{8^{1/3}}{27^{1/3}} = \frac{2}{3}$$ 3) Factorize completely $3x^2 - 27x$. Factor out greatest common factor $3x$: $$3x^2 - 27x = 3x(x - 9)$$ 4) Given the speed-time graph with speed dropping linearly from 9 to 0 in 9 seconds: 4.a) Find the acceleration during the first 8 seconds. Acceleration = change in speed / time: $$a = \frac{v_{final} - v_{initial}}{t} = \frac{0 - 9}{9} = -1 \text{ m/s}^2$$ During first 8 seconds, speed decreases linearly so acceleration is constant: $$a = -1 \text{ m/s}^2$$ 4.b) Find distance covered in first 14 seconds. Speed decreases from 9 to 0 in 9 seconds, then stays at 0 from 9 to 14 seconds (assumed from graph). Distance is area under speed-time graph. Area from 0 to 9 seconds is triangle: $$\text{Area} = \frac{1}{2} \times 9 \times 9 = 40.5 \text{ units}$$ From 9 to 14 seconds speed is 0, so distance = 0. Total distance: $$40.5$$ 4.c) Find average speed if total distance in 14 seconds is 40.5: $$\text{Average speed} = \frac{\text{Distance}}{\text{Time}} = \frac{40.5}{14} \approx 2.89 \text{ m/s}$$ 4.d) Solve the equation $2x + 3 = \frac{1}{6}$. $$2x = \frac{1}{6} - 3 = \frac{1}{6} - \frac{18}{6} = -\frac{17}{6}$$ $$x = -\frac{17}{12}$$ 4.e) Find $\int (6x^2 - 2x + 7) dx$. Integrate term-by-term: $$\int 6x^2 dx = 6 \cdot \frac{x^3}{3} = 2x^3$$ $$\int -2x dx = -2 \cdot \frac{x^2}{2} = -x^2$$ $$\int 7 dx = 7x$$ Therefore: $$\int (6x^2 - 2x + 7) dx = 2x^3 - x^2 + 7x + C$$ 5) Given $f(x) = x - 4$, $h(x) = 2x + 3$: 5.a) Find $h^{-1}(x)$. Let $y = h(x) = 2x + 3$. Solve for $x$: $$y = 2x + 3$$ $$2x = y - 3$$ $$x = \frac{y - 3}{2}$$ So, $$h^{-1}(x) = \frac{x - 3}{2}$$ 5.b) Find $h^{-1}(5)$: $$h^{-1}(5) = \frac{5 - 3}{2} = \frac{2}{2} = 1$$ 5.c) Find $hf(x)$, composition $h(f(x))$: $$hf(x) = h(f(x)) = h(x - 4) = 2(x - 4) + 3 = 2x - 8 + 3 = 2x - 5$$