1. Let $f(x) = x^2 - 1$ and $g(x) = 2x + 3$. Find; (a) $(f \circ g)(x)$ and $(g \circ f)(x)$. Step 1: Calculate $(f \circ g)(x) = f(g(x))$. Substitute $g(x) = 2x + 3$ into $f$: $$f(g(x)) = (2x + 3)^2 - 1 = (2x + 3)(2x + 3) - 1 = 4x^2 + 12x + 9 - 1 = 4x^2 + 12x + 8$$ Step 2: Calculate $(g \circ f)(x) = g(f(x))$. Substitute $f(x) = x^2 - 1$ into $g$: $$g(f(x)) = 2(x^2 - 1) + 3 = 2x^2 - 2 + 3 = 2x^2 + 1$$ (b) Find $(f \circ g)(-1)$ and $(g \circ f)(2)$. Step 3: Calculate $(f \circ g)(-1)$. Using the formula from Step 1: $$(f \circ g)(-1) = 4(-1)^2 + 12(-1) + 8 = 4 - 12 + 8 = 0$$ Step 4: Calculate $(g \circ f)(2)$. Using the formula from Step 2: $$(g \circ f)(2) = 2(2)^2 + 1 = 2(4) + 1 = 8 + 1 = 9$$ 2. Let $f(x) = \frac{x - 3}{x - 2}$ and $g(x) = 3x + b$. Find; (a) $f^{-1}(x)$ and $g^{-1}(x)$. Step 5: Find $f^{-1}(x)$. Let $y = \frac{x - 3}{x - 2}$. To find inverse, swap $x$ and $y$: $$x = \frac{y - 3}{y - 2}$$ Multiply both sides by $(y - 2)$: $$x(y - 2) = y - 3$$ Expand: $$xy - 2x = y - 3$$ Group terms with $y$: $$xy - y = 2x - 3$$ Factor $y$: $$y(x - 1) = 2x - 3$$ Divide both sides: $$y = \frac{2x - 3}{x - 1}$$ So, $$f^{-1}(x) = \frac{2x - 3}{x - 1}$$ Step 6: Find $g^{-1}(x)$. Let $y = 3x + b$. Solve for $x$: $$y = 3x + b \implies 3x = y - b \implies x = \frac{y - b}{3}$$ So, $$g^{-1}(x) = \frac{x - b}{3}$$ (b) Find $f^{-1}(-3)$. Substitute $x = -3$ into $f^{-1}(x)$: $$f^{-1}(-3) = \frac{2(-3) - 3}{-3 - 1} = \frac{-6 - 3}{-4} = \frac{-9}{-4} = \frac{9}{4} = 2.25$$ (c) Given $g^{-1}(5) = 1$, find $b$. Substitute into $g^{-1}(x)$: $$1 = \frac{5 - b}{3} \implies 3 = 5 - b \implies b = 5 - 3 = 2$$ 3. Trajectory $h(t) = -3t^2 + 24t + 36$. (a) Initial height at $t=0$: $$h(0) = -3(0)^2 + 24(0) + 36 = 36$$ (b) Height at $t=2$: $$h(2) = -3(2)^2 + 24(2) + 36 = -3\cdot4 + 48 + 36 = -12 + 48 + 36 = 72$$ (c) Time to reach maximum height (vertex $t$-coordinate of parabola $at^2+bt+c$ is $t= -\frac{b}{2a}$): $$t = -\frac{24}{2 \times (-3)} = -\frac{24}{-6} = 4$$ (d) Maximum height is $h(4)$: $$h(4) = -3(4)^2 + 24(4) + 36 = -3(16) + 96 + 36 = -48 + 96 + 36 = 84$$ 4. Quadratic $f(x) = 3x^2 + 6x + 2$. (a) Express in form $a(x+p)^2 + q$: Step 1: Factor out $3$ from the first two terms: $$f(x) = 3(x^2 + 2x) + 2$$ Step 2: Complete the square inside parentheses: add and subtract $(\frac{2}{2})^2 = 1$: $$3(x^2 + 2x + 1 - 1) + 2 = 3((x + 1)^2 - 1) + 2$$ Step 3: Distribute $3$: $$3(x + 1)^2 - 3 + 2 = 3(x + 1)^2 - 1$$ So, $$f(x) = 3(x + 1)^2 - 1$$ (b) Coordinates of turning point (vertex) are $(-1, -1)$ as seen from vertex form. Because $a = 3 > 0$, the parabola opens upward and the turning point is a minimum. Final answers: (a) $(f \circ g)(x) = 4x^2 + 12x + 8$, $(g \circ f)(x) = 2x^2 + 1$. (b) $(f \circ g)(-1) = 0$, $(g \circ f)(2) = 9$. (c) $f^{-1}(x) = \frac{2x - 3}{x - 1}$, $g^{-1}(x) = \frac{x - b}{3}$. (d) $f^{-1}(-3) = 2.25$. (e) $b = 2$. (f) Initial height $= 36$. (g) Height at 2 sec $= 72$. (h) Time to max height $= 4$ seconds. (i) Maximum height $= 84$ meters. (j) Completed square form $f(x) = 3(x + 1)^2 - 1$. (k) Turning point at $(-1, -1)$ is a minimum.