1. **Problem A:** Sketch the graph of the relation $y=1-x^3$ and find its domain and range. 2. The formula is $y=1-x^3$. This is a cubic function shifted up by 1. 3. The domain of a cubic function is all real numbers because $x$ can take any value. 4. The range is also all real numbers because as $x$ goes to $\pm \infty$, $y$ goes to $\mp \infty$ respectively. 5. To sketch, plot points for values of $x$ such as $-2, -1, 0, 1, 2$: - $x=-2 \Rightarrow y=1-(-2)^3=1+8=9$ - $x=-1 \Rightarrow y=1-(-1)^3=1+1=2$ - $x=0 \Rightarrow y=1-0=1$ - $x=1 \Rightarrow y=1-1=0$ - $x=2 \Rightarrow y=1-8=-7$ 6. The graph passes through these points and has the typical cubic shape flipped vertically and shifted. --- 7. **Problem B:** Sketch the graph of the relation defined by the inequalities $y < -2x + 2$ and $y > x - 3$ and find its domain and range. 8. These inequalities describe the region between two lines: - Line 1: $y = -2x + 2$ - Line 2: $y = x - 3$ 9. The domain is all $x$ values where the two inequalities overlap. 10. Find the intersection point by solving: $$-2x + 2 = x - 3$$ $$-2x - x = -3 - 2$$ $$-3x = -5$$ $$x = \frac{5}{3}$$ 11. Substitute $x=\frac{5}{3}$ into one line to find $y$: $$y = x - 3 = \frac{5}{3} - 3 = \frac{5}{3} - \frac{9}{3} = -\frac{4}{3}$$ 12. The intersection point is $\left(\frac{5}{3}, -\frac{4}{3}\right)$. 13. The region satisfying both inequalities is the area between these lines below $y=-2x+2$ and above $y=x-3$. 14. Since the lines extend infinitely, the domain is all real numbers. 15. The range is all $y$ values between the two lines, which depends on $x$ but is bounded between these lines. --- **Final answers:** - Problem A: Domain is $\mathbb{R}$, Range is $\mathbb{R}$. - Problem B: Domain is $\mathbb{R}$, Range is the set of $y$ such that $x-3 < y < -2x+2$ for all $x$ in $\mathbb{R}$.