Answer :
We are given the functions
[tex]$$
h(x)=\frac{3}{3-2x} \quad \text{and} \quad l(x)=\frac{2x}{2x-3}.
$$[/tex]
We will determine the domains for each function and then simplify and find the domains of the combinations as required.
──────────────────────────────
1. Domains of [tex]$h(x)$[/tex] and [tex]$l(x)$[/tex]
For a rational function, the domain consists of all real numbers except where the denominator is zero.
• For [tex]$h(x)$[/tex], the denominator is [tex]$3-2x$[/tex]. Set it equal to zero to find the excluded value:
[tex]$$
3-2x=0 \quad \Longrightarrow \quad x=\frac{3}{2}.
$$[/tex]
Thus,
[tex]$$
D_h = \{x \in \mathbb{R} \mid x \neq \frac{3}{2}\}.
$$[/tex]
• For [tex]$l(x)$[/tex], the denominator is [tex]$2x-3$[/tex]. Setting it to zero gives:
[tex]$$
2x-3=0 \quad \Longrightarrow \quad x=\frac{3}{2}.
$$[/tex]
Thus,
[tex]$$
D_l = \{x \in \mathbb{R} \mid x \neq \frac{3}{2}\}.
$$[/tex]
──────────────────────────────
2. Simplify [tex]$h(x)+l(x)$[/tex] and determine [tex]$D_{h+l}$[/tex]
We have:
[tex]$$
h(x)+l(x)=\frac{3}{3-2x}+\frac{2x}{2x-3}.
$$[/tex]
Notice that
[tex]$$
2x-3 = -(3-2x).
$$[/tex]
So we can rewrite
[tex]$$
l(x)=\frac{2x}{2x-3}=-\frac{2x}{3-2x}.
$$[/tex]
Now,
[tex]$$
h(x) + l(x) = \frac{3}{3-2x} - \frac{2x}{3-2x}
=\frac{3-2x}{3-2x}=1,
$$[/tex]
provided that [tex]$3-2x\neq0$[/tex], i.e. [tex]$x\neq \frac{3}{2}$[/tex].
Thus,
[tex]$$
h(x)+l(x)=1 \quad \text{with domain} \quad D_{h+l} = \{x \in \mathbb{R} \mid x \neq \frac{3}{2}\}.
$$[/tex]
──────────────────────────────
3. Simplify [tex]$h(x)-l(x)$[/tex] and determine [tex]$D_{h-l}$[/tex]
Similarly, we compute:
[tex]$$
h(x)-l(x)=\frac{3}{3-2x}-\frac{2x}{2x-3}.
$$[/tex]
Again, use [tex]$2x-3=-(3-2x)$[/tex] to write
[tex]$$
l(x)=-\frac{2x}{3-2x}.
$$[/tex]
Then,
[tex]$$
h(x)-l(x)=\frac{3}{3-2x}-\left(-\frac{2x}{3-2x}\right)
=\frac{3+2x}{3-2x}.
$$[/tex]
The only restriction comes from the denominator [tex]$3-2x \neq 0$[/tex], so
[tex]$$
D_{h-l} = \{x \in \mathbb{R} \mid x \neq \frac{3}{2}\}.
$$[/tex]
──────────────────────────────
4. Simplify [tex]$h(x)\cdot l(x)$[/tex] and determine [tex]$D_{h\cdot l}$[/tex]
We have:
[tex]$$
h(x)\cdot l(x)=\frac{3}{3-2x}\cdot \frac{2x}{2x-3}.
$$[/tex]
Again, rewriting [tex]$2x-3=-(3-2x)$[/tex], we get:
[tex]$$
l(x)= -\frac{2x}{3-2x}.
$$[/tex]
Thus,
[tex]$$
h(x)\cdot l(x)=\frac{3}{3-2x}\cdot \left(-\frac{2x}{3-2x}\right)
=-\frac{6x}{(3-2x)^2}.
$$[/tex]
The domain remains restricted by [tex]$3-2x \neq 0$[/tex], so
[tex]$$
D_{h\cdot l} = \{x \in \mathbb{R} \mid x \neq \frac{3}{2}\}.
$$[/tex]
──────────────────────────────
5. Simplify [tex]$\dfrac{h(x)}{l(x)}$[/tex] and determine [tex]$D_{\frac{h}{l}}$[/tex]
We compute:
[tex]$$
\frac{h(x)}{l(x)}=\frac{\frac{3}{3-2x}}{\frac{2x}{2x-3}}=\frac{3}{3-2x}\cdot \frac{2x-3}{2x}.
$$[/tex]
Using [tex]$2x-3=-(3-2x)$[/tex] gives:
[tex]$$
\frac{h(x)}{l(x)}=\frac{3}{3-2x}\cdot \frac{-(3-2x)}{2x}
=-\frac{3}{2x}.
$$[/tex]
Now, besides the restriction [tex]$x\neq \frac{3}{2}$[/tex] coming from the original domains of [tex]$h(x)$[/tex] and [tex]$l(x)$[/tex], we must also have [tex]$2x\neq0$[/tex], which means [tex]$x\neq 0$[/tex]. Thus,
[tex]$$
D_{\frac{h}{l}} = \{x \in \mathbb{R} \mid x \neq \frac{3}{2} \text{ and } x \neq 0\}.
$$[/tex]
──────────────────────────────
Summary of Answers
1. Domain of [tex]$h(x)$[/tex]:
[tex]$$
D_h = \{x\in\mathbb{R} : x\neq \frac{3}{2}\},
$$[/tex]
Domain of [tex]$l(x)$[/tex]:
[tex]$$
D_l = \{x\in\mathbb{R} : x\neq \frac{3}{2}\}.
$$[/tex]
2. For [tex]$h(x)+l(x)$[/tex]:
[tex]$$
h(x)+l(x)=1, \quad D_{h+l}=\{x\in\mathbb{R}: x\neq \frac{3}{2}\}.
$$[/tex]
3. For [tex]$h(x)-l(x)$[/tex]:
[tex]$$
h(x)-l(x)=\frac{3+2x}{3-2x}, \quad D_{h-l}=\{x\in\mathbb{R}: x\neq \frac{3}{2}\}.
$$[/tex]
4. For [tex]$h(x)\cdot l(x)$[/tex]:
[tex]$$
h(x)\cdot l(x)=-\frac{6x}{(3-2x)^2}, \quad D_{h\cdot l}=\{x\in\mathbb{R}: x\neq \frac{3}{2}\}.
$$[/tex]
5. For [tex]$\dfrac{h(x)}{l(x)}$[/tex]:
[tex]$$
\frac{h(x)}{l(x)}=-\frac{3}{2x}, \quad D_{\frac{h}{l}} = \{x\in\mathbb{R}: x\neq \frac{3}{2}\text{ and }x\neq 0\}.
$$[/tex]
Each step ensures that the restrictions from the denominators are maintained.
[tex]$$
h(x)=\frac{3}{3-2x} \quad \text{and} \quad l(x)=\frac{2x}{2x-3}.
$$[/tex]
We will determine the domains for each function and then simplify and find the domains of the combinations as required.
──────────────────────────────
1. Domains of [tex]$h(x)$[/tex] and [tex]$l(x)$[/tex]
For a rational function, the domain consists of all real numbers except where the denominator is zero.
• For [tex]$h(x)$[/tex], the denominator is [tex]$3-2x$[/tex]. Set it equal to zero to find the excluded value:
[tex]$$
3-2x=0 \quad \Longrightarrow \quad x=\frac{3}{2}.
$$[/tex]
Thus,
[tex]$$
D_h = \{x \in \mathbb{R} \mid x \neq \frac{3}{2}\}.
$$[/tex]
• For [tex]$l(x)$[/tex], the denominator is [tex]$2x-3$[/tex]. Setting it to zero gives:
[tex]$$
2x-3=0 \quad \Longrightarrow \quad x=\frac{3}{2}.
$$[/tex]
Thus,
[tex]$$
D_l = \{x \in \mathbb{R} \mid x \neq \frac{3}{2}\}.
$$[/tex]
──────────────────────────────
2. Simplify [tex]$h(x)+l(x)$[/tex] and determine [tex]$D_{h+l}$[/tex]
We have:
[tex]$$
h(x)+l(x)=\frac{3}{3-2x}+\frac{2x}{2x-3}.
$$[/tex]
Notice that
[tex]$$
2x-3 = -(3-2x).
$$[/tex]
So we can rewrite
[tex]$$
l(x)=\frac{2x}{2x-3}=-\frac{2x}{3-2x}.
$$[/tex]
Now,
[tex]$$
h(x) + l(x) = \frac{3}{3-2x} - \frac{2x}{3-2x}
=\frac{3-2x}{3-2x}=1,
$$[/tex]
provided that [tex]$3-2x\neq0$[/tex], i.e. [tex]$x\neq \frac{3}{2}$[/tex].
Thus,
[tex]$$
h(x)+l(x)=1 \quad \text{with domain} \quad D_{h+l} = \{x \in \mathbb{R} \mid x \neq \frac{3}{2}\}.
$$[/tex]
──────────────────────────────
3. Simplify [tex]$h(x)-l(x)$[/tex] and determine [tex]$D_{h-l}$[/tex]
Similarly, we compute:
[tex]$$
h(x)-l(x)=\frac{3}{3-2x}-\frac{2x}{2x-3}.
$$[/tex]
Again, use [tex]$2x-3=-(3-2x)$[/tex] to write
[tex]$$
l(x)=-\frac{2x}{3-2x}.
$$[/tex]
Then,
[tex]$$
h(x)-l(x)=\frac{3}{3-2x}-\left(-\frac{2x}{3-2x}\right)
=\frac{3+2x}{3-2x}.
$$[/tex]
The only restriction comes from the denominator [tex]$3-2x \neq 0$[/tex], so
[tex]$$
D_{h-l} = \{x \in \mathbb{R} \mid x \neq \frac{3}{2}\}.
$$[/tex]
──────────────────────────────
4. Simplify [tex]$h(x)\cdot l(x)$[/tex] and determine [tex]$D_{h\cdot l}$[/tex]
We have:
[tex]$$
h(x)\cdot l(x)=\frac{3}{3-2x}\cdot \frac{2x}{2x-3}.
$$[/tex]
Again, rewriting [tex]$2x-3=-(3-2x)$[/tex], we get:
[tex]$$
l(x)= -\frac{2x}{3-2x}.
$$[/tex]
Thus,
[tex]$$
h(x)\cdot l(x)=\frac{3}{3-2x}\cdot \left(-\frac{2x}{3-2x}\right)
=-\frac{6x}{(3-2x)^2}.
$$[/tex]
The domain remains restricted by [tex]$3-2x \neq 0$[/tex], so
[tex]$$
D_{h\cdot l} = \{x \in \mathbb{R} \mid x \neq \frac{3}{2}\}.
$$[/tex]
──────────────────────────────
5. Simplify [tex]$\dfrac{h(x)}{l(x)}$[/tex] and determine [tex]$D_{\frac{h}{l}}$[/tex]
We compute:
[tex]$$
\frac{h(x)}{l(x)}=\frac{\frac{3}{3-2x}}{\frac{2x}{2x-3}}=\frac{3}{3-2x}\cdot \frac{2x-3}{2x}.
$$[/tex]
Using [tex]$2x-3=-(3-2x)$[/tex] gives:
[tex]$$
\frac{h(x)}{l(x)}=\frac{3}{3-2x}\cdot \frac{-(3-2x)}{2x}
=-\frac{3}{2x}.
$$[/tex]
Now, besides the restriction [tex]$x\neq \frac{3}{2}$[/tex] coming from the original domains of [tex]$h(x)$[/tex] and [tex]$l(x)$[/tex], we must also have [tex]$2x\neq0$[/tex], which means [tex]$x\neq 0$[/tex]. Thus,
[tex]$$
D_{\frac{h}{l}} = \{x \in \mathbb{R} \mid x \neq \frac{3}{2} \text{ and } x \neq 0\}.
$$[/tex]
──────────────────────────────
Summary of Answers
1. Domain of [tex]$h(x)$[/tex]:
[tex]$$
D_h = \{x\in\mathbb{R} : x\neq \frac{3}{2}\},
$$[/tex]
Domain of [tex]$l(x)$[/tex]:
[tex]$$
D_l = \{x\in\mathbb{R} : x\neq \frac{3}{2}\}.
$$[/tex]
2. For [tex]$h(x)+l(x)$[/tex]:
[tex]$$
h(x)+l(x)=1, \quad D_{h+l}=\{x\in\mathbb{R}: x\neq \frac{3}{2}\}.
$$[/tex]
3. For [tex]$h(x)-l(x)$[/tex]:
[tex]$$
h(x)-l(x)=\frac{3+2x}{3-2x}, \quad D_{h-l}=\{x\in\mathbb{R}: x\neq \frac{3}{2}\}.
$$[/tex]
4. For [tex]$h(x)\cdot l(x)$[/tex]:
[tex]$$
h(x)\cdot l(x)=-\frac{6x}{(3-2x)^2}, \quad D_{h\cdot l}=\{x\in\mathbb{R}: x\neq \frac{3}{2}\}.
$$[/tex]
5. For [tex]$\dfrac{h(x)}{l(x)}$[/tex]:
[tex]$$
\frac{h(x)}{l(x)}=-\frac{3}{2x}, \quad D_{\frac{h}{l}} = \{x\in\mathbb{R}: x\neq \frac{3}{2}\text{ and }x\neq 0\}.
$$[/tex]
Each step ensures that the restrictions from the denominators are maintained.
