Answer:
[tex]\textsf{(a)} \quad -4 < x\leq 1[/tex]
[tex]\textsf{(b)} \quad [3, 19)[/tex]
Step-by-step explanation:
Part (a)
Given compound inequality:
[tex]\dfrac{9+x}{5} < 5+x\leq 6[/tex]
[tex]\textsf{If\: $a < u\leq b$ \:then\: $a < u$ \:and\: $u\leq b$}.[/tex]
[tex]\textsf{Therefore \:$\dfrac{9+x}{5} < 5+x$ \:and\: $5+x\leq 6$}[/tex]
Solve the inequalities separately:
Inequality 1
[tex]\begin{aligned}\dfrac{9+x}{5} & < 5+x\\9+x & < 5(5+x)\\9+x & < 25+5x\\9+x -5x& < 25+5x-5x\\9-4x & < 25\\ 9-4x-9 & < 25-9\\-4x & < 16\\-x & < 4\\x & > -4\\ \end{aligned}[/tex]
Inequality 2
[tex]\begin{aligned} 5+x & \leq 6\\ 5+x-5 & \leq 6-5\\x & \leq 1 \end{aligned}[/tex]
Combine the intervals:
[tex]-4 < x\leq 1[/tex]
Part (b)
Given equation:
[tex]y=x^2+3[/tex]
As the domain is restricted to -4 < x ≤ 1, the range is also restricted.
The vertex (minimum point) of [tex]y = x^2 + 3[/tex] is when x = 0.
As x = 0 lies within the restricted domain, y = 3 is the lowest value of the range.
[tex]x=-4 \implies y=(-4)^2+3=19[/tex]
Therefore, the range is:
- Solution: 3 ≤ y < 19.
- Interval notation: [3, 19)
