1. Problem statement: Given graphs of functions $f$ and $g$, answer parts a through e based on the graph descriptions. 2. (a) Find $f(-4)$ and $g(3)$: - From the description, $f$ starts below the $x$-axis at $x = -4$, so $f(-4) < 0$. Exact value isn't given but visually estimated near $-1$. - $g(3)$ is described as rising steeply above $y=2$ at $x=3$, so $g(3) > 2$, estimated near $2.5$. 3. (b) Find $x$ where $f(x) = g(x)$: - The curves intersect between $0$ and $2$ once (where $g$ crosses $f$), so $f(x) = g(x)$ near some $x e 0$ in $(0,2)$. 4. (c) Estimate solution of $f(x) = -1$: - $f(x) = -1$ occurs where $f$ passes $y=-1$. From graph, near $x = -4$ and also after $x=2$ where $f$ falls below $0$. Estimating near $x = -4$ and a value greater than $2$. 5. (d) Domain and range of $f$: - Domain: $f$ is graphed from $x=-4$ and extends past $x=2$, so domain approximately $[-4, ext{beyond } 2]$ (assuming continuous). - Range: $f$ peaks just above $2$ and goes below $0$, so range approximately $(- ext{some negative}, >2)$. 6. (e) Domain and range of $g$: - Domain includes at least $[-4,3]$ since points at $-4,0,3$ are described. - Range: starts above $0$ at $x=-4$, dips to about $0$ near $x=0$, rises above $2$ at $x=3$. So range approximately $[0, >2.5]$. --- 7. Problem 05 Find domain of $f(x) = \frac{5x - 2}{x - \sqrt{2x + 3}}$: - Denominator $\neq 0$: Solve $x - \sqrt{2x + 3} \neq 0$. - Also under root $2x + 3 \geq 0 \implies x \geq -\frac{3}{2}$. - Set denominator $=0$ gives $x=\sqrt{2x + 3}$. - Square both sides: $x^2 = 2x + 3 \Rightarrow x^2 - 2x -3=0$. - Solve quadratic: $x = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm 4}{2}$. - Roots: $x=3$ or $x=-1$, test in original equation for validity. - Substitute back: For $x=3$, RHS $=\sqrt{9} = 3$, so denominator $= 3-3=0$. - For $x=-1$, RHS $=\sqrt{1} =1$, denominator $= -1 - 1 = -2 \neq 0$, so exclude $x=3$. - Domain: $x \geq -\frac{3}{2}$, $x \neq 3$. 8. Problem 07 find inverses: (a) $f(x) = x^2 + 4$, $x \geq 4$. - $y = x^2 + 4 \Rightarrow x^2 = y -4$, use +ve root due to $x \geq 4$. - Inverse: $f^{-1}(y) = \sqrt{y - 4}$. - Domain of $f$: $[4, \infty)$, range $[4^2 + 4=20, \infty)$. - Inverse domain: $[20, \infty)$, range $[4, \infty)$. (b) $f(x) = 3 - 2x$. - Solve $y=3-2x$ for $x$: $x = \frac{3 - y}{2}$. - $f^{-1}(y) = \frac{3 - y}{2}$. - Domain and range: All real numbers. (c) $g(x)=\frac{2 - x}{3 + x}$. - Set $y=\frac{2 - x}{3 + x}$, solve for $x$: $$ y(3 + x) = 2 - x \Rightarrow 3y + yx = 2 - x $$ $$ yx + x = 2 - 3y \Rightarrow x(y + 1) = 2 - 3y $$ $$ x = \frac{2 - 3y}{y + 1}, y \neq -1 $$ - Inverse: $g^{-1}(y) = \frac{2 - 3y}{y + 1}$. - Domain excludes $x=-3$, range excludes $y=-1$. 9. Problem 01 Homework difference quotients examples have multiple parts; total distinct questions here are multiple. 10. For all other questions, similar detailed stepwise explanations can be provided if requested each separately. Final answers summarized for main given graph problem: $f(-4) \approx -1$ $g(3) \approx 2.5$ $f(x) = g(x)$ near $x$ between $0$ and $2$ $ f(x) = -1$ near $x = -4$ and just above $2$ $\text{Domain}_f \approx [-4, >2]$, $\text{Range}_f \approx (-\infty, >2]$ $\text{Domain}_g \approx [-4, 3]$, $\text{Range}_g \approx [0, >2.5]$