1. The problem asks if the function $B(t)$, representing temperature in Boston over time $t$ (hours after midnight), is one-to-one. 2. A function is one-to-one if each output corresponds to exactly one input. This means the function passes the Horizontal Line Test: no horizontal line intersects the graph more than once. 3. Since temperature can repeat at different times (e.g., morning and evening), $B(t)$ is generally not one-to-one unless specified otherwise. 4. For the graph of $y=f(x)$ on $-6 \leq x \leq 6$, to check if $f$ is one-to-one, apply the Horizontal Line Test: if any horizontal line crosses the graph more than once, $f$ is not one-to-one. 5. The graph has three segments: increasing, sharply rising, then descending. Because it descends after rising, some horizontal lines will intersect the graph twice. 6. Therefore, $f$ is not one-to-one on its domain. 7. Find the inverse of $y=\sqrt{x}+3$. 8. Start by writing $y=\sqrt{x}+3$. 9. Swap $x$ and $y$ to find the inverse: $x=\sqrt{y}+3$. 10. Solve for $y$: $$x-3=\sqrt{y}$$ $$y=(x-3)^2$$ 11. The domain of the original function is $x \geq 0$ (since square root is defined for non-negative $x$). 12. The range of the original function is $y \geq 3$ (since $\sqrt{x} \geq 0$). 13. For the inverse, domain and range swap: - Domain of $f^{-1}$ is $x \geq 3$. - Range of $f^{-1}$ is $y \geq 0$. 14. Let $f(x) = \frac{1}{x+5}$. 15. Find the inverse $f^{-1}$: $$y=\frac{1}{x+5}$$ Swap $x$ and $y$: $$x=\frac{1}{y+5}$$ Solve for $y$: $$x(y+5)=1$$ $$xy+5x=1$$ $$xy=1-5x$$ $$y=\frac{1-5x}{x}$$ 16. Domain of $f$ is all real numbers except $x=-5$ (denominator zero). 17. Range of $f$ is all real numbers except $0$ (since $\frac{1}{x+5}$ never equals zero). 18. Domain of $f^{-1}$ is range of $f$, so all real numbers except $0$. 19. Range of $f^{-1}$ is domain of $f$, all real numbers except $-5$. 20. It is possible to find domain and range of $f^{-1}$ without explicit formula by swapping domain and range of $f$. 21. For function $g$ with vertical asymptote at $x=-5$ and given values: | x | g(x) | |---|-------| | -5 | undefined | | -2 | -14 | | 0 | -11 | | 3 | -1 | | 6 | 0 | 22. a. The y-intercept of $g^{-1}$ is the $y$-value when $x=0$ for $g^{-1}$, which corresponds to $x=0$ in $g$. 23. Since $g(0)=-11$, the point $(0,-11)$ is on $g$, so $( -11, 0)$ is on $g^{-1}$. 24. Therefore, the y-intercept of $g^{-1}$ is $-11$. 25. b. The horizontal asymptote of $g$ corresponds to the vertical asymptote of $g^{-1}$. 26. Since $g$ has vertical asymptote at $x=-5$, $g^{-1}$ has horizontal asymptote at $y=-5$. Final answers: - $B(t)$ is not one-to-one. - $f$ is not one-to-one. - Inverse of $y=\sqrt{x}+3$ is $y=(x-3)^2$ with domain $x \geq 3$ and range $y \geq 0$. - Domain of $f^{-1}$ is all real numbers except $0$, range is all real numbers except $-5$. - Y-intercept of $g^{-1}$ is $-11$. - Horizontal asymptote of $g^{-1}$ is $y=-5$.