1. Define the concepts: (i) Domain is the set of all possible input values (x-values) for which the function is defined. (ii) Range is the set of all possible output values (y-values) that the function can take. 2. Find domain and range: (i) For $y=24+10x-x^2$: - This is a quadratic function (a parabola) opening downward because the coefficient of $x^2$ is negative. - Domain: All real numbers, $\mathbf{\mathbb{R}}$, since any $x$ can be plugged in. - Range: Maximum value at vertex. The vertex $x$-coordinate is $x=-\frac{b}{2a}=-\frac{10}{2(-1)}=5$. - Substitute $x=5$ into $y$: $$y=24+10(5)-5^2=24+50-25=49$$ - So, range is $(-\infty, 49]$. (ii) For $y=\frac{1}{6\sqrt{x^2}}$: - $\sqrt{x^2}=|x|$. - Function becomes $y=\frac{1}{6|x|}$. - Domain: $x\neq 0$ because denominator can't be zero. - Range: Since $|x|>0$, $y>0$ and as $|x|\to \infty$, $y\to 0^{+}$. - So, range is $(0, \infty)$. 3. Find inverse functions: (i) For $f:x\to \frac{x+1}{x+2}$: - Set $y=\frac{x+1}{x+2}$. - Solve for $x$: $$y(x+2) = x+1 \Rightarrow yx + 2y = x+1 \Rightarrow yx - x = 1 - 2y \Rightarrow x(y-1) = 1-2y \Rightarrow x = \frac{1-2y}{y-1}$$ - Thus, inverse $f^{-1}: y \to \frac{1-2y}{y-1}$. (ii) For $g:y \to 3y + 2$: - Set $z=3y+2$. - Solve for $y$: $$3y = z-2 \Rightarrow y=\frac{z-2}{3}$$ - Thus, inverse $g^{-1}: z \to \frac{z-2}{3}$. (iii) Find $f^{-1}(g^{-1}(2))$: - Compute $g^{-1}(2)$: $$g^{-1}(2) = \frac{2-2}{3} = 0$$ - Compute $f^{-1}(0)$: $$f^{-1}(0) = \frac{1-2(0)}{0-1} = \frac{1}{-1} = -1$$ Final answers: (i) Domain and Range: - $y=24+10x-x^2$ domain: $\mathbf{\mathbb{R}}$, range: $(-\infty, 49]$. - $y=\frac{1}{6\sqrt{x^2}}$ domain: $\mathbf{\mathbb{R}} \setminus \{0\}$, range: $(0, \infty)$. (ii) Inverses: - $f^{-1}: y \to \frac{1-2y}{y-1}$. - $g^{-1}: z \to \frac{z-2}{3}$. (iii) $f^{-1}(g^{-1}(2)) = -1$.