Answer :
We are given the functions
[tex]$$
h(x)=\frac{3}{3-2x} \quad \text{and} \quad l(x)=\frac{2x}{2x-3}.
$$[/tex]
Let us go through the steps to determine the domains and the results for each operation.
─────────────────────────────
Step 1. Determining the Domains of [tex]\(h(x)\)[/tex] and [tex]\(l(x)\)[/tex]
For [tex]\(h(x)=\frac{3}{3-2x}\)[/tex], the denominator must not be zero:
[tex]\[
3-2x \neq 0 \quad\Longrightarrow\quad x \neq \frac{3}{2}.
\][/tex]
Thus, the domain of [tex]\(h\)[/tex] is
[tex]\[
D_h=\{x\in\mathbb{R} : x\neq \frac{3}{2}\}.
\][/tex]
For [tex]\(l(x)=\frac{2x}{2x-3}\)[/tex], we require
[tex]\[
2x-3 \neq 0 \quad\Longrightarrow\quad x\neq \frac{3}{2}.
\][/tex]
So, the domain of [tex]\(l\)[/tex] is
[tex]\[
D_l=\{x\in\mathbb{R} : x\neq \frac{3}{2}\}.
\][/tex]
─────────────────────────────
Step 2. Calculating [tex]\(h(x)+l(x)\)[/tex] and Its Domain
Notice that in [tex]\(l(x)\)[/tex], we have
[tex]\[
2x-3=-(3-2x).
\][/tex]
So, we may rewrite
[tex]\[
l(x) = \frac{2x}{2x-3} = -\frac{2x}{3-2x}.
\][/tex]
Then,
[tex]\[
h(x)+l(x) = \frac{3}{3-2x} - \frac{2x}{3-2x} = \frac{3-2x}{3-2x}.
\][/tex]
For [tex]\(x\neq \frac{3}{2}\)[/tex] (since [tex]\(3-2x\neq 0\)[/tex]), the fraction in the numerator simplifies to
[tex]\[
h(x)+l(x)=1.
\][/tex]
The domain of [tex]\(h+l\)[/tex] is the common domain of [tex]\(h\)[/tex] and [tex]\(l\)[/tex]:
[tex]\[
D_{h+l}=\{x\in\mathbb{R} : x\neq \frac{3}{2}\}.
\][/tex]
─────────────────────────────
Step 3. Calculating [tex]\(h(x)-l(x)\)[/tex] and Its Domain
Using the same rewriting for [tex]\(l(x)\)[/tex]:
[tex]\[
h(x)-l(x)=\frac{3}{3-2x} + \frac{2x}{3-2x} = \frac{3+2x}{3-2x}.
\][/tex]
The domain is again
[tex]\[
D_{h-l}=\{x \in\mathbb{R} : x\neq \frac{3}{2}\}.
\][/tex]
─────────────────────────────
Step 4. Calculating [tex]\(h(x)\cdot l(x)\)[/tex] and Its Domain
Multiplying the two functions we have
[tex]\[
h(x)\cdot l(x)=\frac{3}{3-2x}\cdot \frac{2x}{2x-3}.
\][/tex]
Notice that
[tex]\[
(3-2x)(2x-3) = (3-2x)\Big[-(3-2x)\Big] = -\bigl(3-2x\bigr)^2.
\][/tex]
Thus, we can simplify as follows:
[tex]\[
h(x) \cdot l(x) = \frac{6x}{-(3-2x)^2}=-\frac{6x}{(3-2x)^2}.
\][/tex]
The domain remains the common domain of both functions:
[tex]\[
D_{h\cdot l}=\{x\in\mathbb{R} : x\neq \frac{3}{2}\}.
\][/tex]
─────────────────────────────
Step 5. Calculating [tex]\(\frac{h(x)}{l(x)}\)[/tex] and Its Domain
We have
[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]
Again, using [tex]\(2x-3=-(3-2x)\)[/tex], we get:
[tex]\[
\frac{h(x)}{l(x)}=\frac{3}{3-2x}\cdot\frac{-\left(3-2x\right)}{2x}=-\frac{3}{2x}.
\][/tex]
In addition to the restriction [tex]\(x\neq \frac{3}{2}\)[/tex] (from the domains of [tex]\(h\)[/tex] and [tex]\(l\)[/tex]), we must also avoid values that make the denominator of this final expression zero (i.e. [tex]\(2x\neq 0\)[/tex] or [tex]\(x\neq 0\)[/tex]). Therefore, the domain is
[tex]\[
D_{h/l}=\{x\in\mathbb{R}: x\neq \frac{3}{2} \text{ and } x\neq 0\}.
\][/tex]
─────────────────────────────
Summary of Answers
1. Domains:
- [tex]\( D_h=\{x\in\mathbb{R} : x\neq \frac{3}{2}\} \)[/tex]
- [tex]\( D_l=\{x\in\mathbb{R} : x\neq \frac{3}{2}\} \)[/tex]
2. For [tex]\(h(x)+l(x)\)[/tex]:
- Expression: [tex]\( h(x)+l(x)=1 \)[/tex]
- Domain: [tex]\( D_{h+l}=\{x\in\mathbb{R} : x\neq \frac{3}{2}\} \)[/tex]
3. For [tex]\(h(x)-l(x)\)[/tex]:
- Expression: [tex]\( h(x)-l(x)=\frac{3+2x}{3-2x} \)[/tex]
- Domain: [tex]\( D_{h-l}=\{x\in\mathbb{R} : x\neq \frac{3}{2}\} \)[/tex]
4. For [tex]\(h(x)\cdot l(x)\)[/tex]:
- Expression: [tex]\( h(x)\cdot l(x)=-\frac{6x}{(3-2x)^2} \)[/tex]
- Domain: [tex]\( D_{h\cdot l}=\{x\in\mathbb{R} : x\neq \frac{3}{2}\} \)[/tex]
5. For [tex]\(\frac{h(x)}{l(x)}\)[/tex]:
- Expression: [tex]\( \frac{h(x)}{l(x)}=-\frac{3}{2x} \)[/tex]
- Domain: [tex]\( D_{h/l}=\{x\in\mathbb{R} : x\neq \frac{3}{2} \text{ and } x\neq 0\} \)[/tex]
Each step carefully considers the conditions under which the original functions and their combinations are defined.
[tex]$$
h(x)=\frac{3}{3-2x} \quad \text{and} \quad l(x)=\frac{2x}{2x-3}.
$$[/tex]
Let us go through the steps to determine the domains and the results for each operation.
─────────────────────────────
Step 1. Determining the Domains of [tex]\(h(x)\)[/tex] and [tex]\(l(x)\)[/tex]
For [tex]\(h(x)=\frac{3}{3-2x}\)[/tex], the denominator must not be zero:
[tex]\[
3-2x \neq 0 \quad\Longrightarrow\quad x \neq \frac{3}{2}.
\][/tex]
Thus, the domain of [tex]\(h\)[/tex] is
[tex]\[
D_h=\{x\in\mathbb{R} : x\neq \frac{3}{2}\}.
\][/tex]
For [tex]\(l(x)=\frac{2x}{2x-3}\)[/tex], we require
[tex]\[
2x-3 \neq 0 \quad\Longrightarrow\quad x\neq \frac{3}{2}.
\][/tex]
So, the domain of [tex]\(l\)[/tex] is
[tex]\[
D_l=\{x\in\mathbb{R} : x\neq \frac{3}{2}\}.
\][/tex]
─────────────────────────────
Step 2. Calculating [tex]\(h(x)+l(x)\)[/tex] and Its Domain
Notice that in [tex]\(l(x)\)[/tex], we have
[tex]\[
2x-3=-(3-2x).
\][/tex]
So, we may rewrite
[tex]\[
l(x) = \frac{2x}{2x-3} = -\frac{2x}{3-2x}.
\][/tex]
Then,
[tex]\[
h(x)+l(x) = \frac{3}{3-2x} - \frac{2x}{3-2x} = \frac{3-2x}{3-2x}.
\][/tex]
For [tex]\(x\neq \frac{3}{2}\)[/tex] (since [tex]\(3-2x\neq 0\)[/tex]), the fraction in the numerator simplifies to
[tex]\[
h(x)+l(x)=1.
\][/tex]
The domain of [tex]\(h+l\)[/tex] is the common domain of [tex]\(h\)[/tex] and [tex]\(l\)[/tex]:
[tex]\[
D_{h+l}=\{x\in\mathbb{R} : x\neq \frac{3}{2}\}.
\][/tex]
─────────────────────────────
Step 3. Calculating [tex]\(h(x)-l(x)\)[/tex] and Its Domain
Using the same rewriting for [tex]\(l(x)\)[/tex]:
[tex]\[
h(x)-l(x)=\frac{3}{3-2x} + \frac{2x}{3-2x} = \frac{3+2x}{3-2x}.
\][/tex]
The domain is again
[tex]\[
D_{h-l}=\{x \in\mathbb{R} : x\neq \frac{3}{2}\}.
\][/tex]
─────────────────────────────
Step 4. Calculating [tex]\(h(x)\cdot l(x)\)[/tex] and Its Domain
Multiplying the two functions we have
[tex]\[
h(x)\cdot l(x)=\frac{3}{3-2x}\cdot \frac{2x}{2x-3}.
\][/tex]
Notice that
[tex]\[
(3-2x)(2x-3) = (3-2x)\Big[-(3-2x)\Big] = -\bigl(3-2x\bigr)^2.
\][/tex]
Thus, we can simplify as follows:
[tex]\[
h(x) \cdot l(x) = \frac{6x}{-(3-2x)^2}=-\frac{6x}{(3-2x)^2}.
\][/tex]
The domain remains the common domain of both functions:
[tex]\[
D_{h\cdot l}=\{x\in\mathbb{R} : x\neq \frac{3}{2}\}.
\][/tex]
─────────────────────────────
Step 5. Calculating [tex]\(\frac{h(x)}{l(x)}\)[/tex] and Its Domain
We have
[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]
Again, using [tex]\(2x-3=-(3-2x)\)[/tex], we get:
[tex]\[
\frac{h(x)}{l(x)}=\frac{3}{3-2x}\cdot\frac{-\left(3-2x\right)}{2x}=-\frac{3}{2x}.
\][/tex]
In addition to the restriction [tex]\(x\neq \frac{3}{2}\)[/tex] (from the domains of [tex]\(h\)[/tex] and [tex]\(l\)[/tex]), we must also avoid values that make the denominator of this final expression zero (i.e. [tex]\(2x\neq 0\)[/tex] or [tex]\(x\neq 0\)[/tex]). Therefore, the domain is
[tex]\[
D_{h/l}=\{x\in\mathbb{R}: x\neq \frac{3}{2} \text{ and } x\neq 0\}.
\][/tex]
─────────────────────────────
Summary of Answers
1. Domains:
- [tex]\( D_h=\{x\in\mathbb{R} : x\neq \frac{3}{2}\} \)[/tex]
- [tex]\( D_l=\{x\in\mathbb{R} : x\neq \frac{3}{2}\} \)[/tex]
2. For [tex]\(h(x)+l(x)\)[/tex]:
- Expression: [tex]\( h(x)+l(x)=1 \)[/tex]
- Domain: [tex]\( D_{h+l}=\{x\in\mathbb{R} : x\neq \frac{3}{2}\} \)[/tex]
3. For [tex]\(h(x)-l(x)\)[/tex]:
- Expression: [tex]\( h(x)-l(x)=\frac{3+2x}{3-2x} \)[/tex]
- Domain: [tex]\( D_{h-l}=\{x\in\mathbb{R} : x\neq \frac{3}{2}\} \)[/tex]
4. For [tex]\(h(x)\cdot l(x)\)[/tex]:
- Expression: [tex]\( h(x)\cdot l(x)=-\frac{6x}{(3-2x)^2} \)[/tex]
- Domain: [tex]\( D_{h\cdot l}=\{x\in\mathbb{R} : x\neq \frac{3}{2}\} \)[/tex]
5. For [tex]\(\frac{h(x)}{l(x)}\)[/tex]:
- Expression: [tex]\( \frac{h(x)}{l(x)}=-\frac{3}{2x} \)[/tex]
- Domain: [tex]\( D_{h/l}=\{x\in\mathbb{R} : x\neq \frac{3}{2} \text{ and } x\neq 0\} \)[/tex]
Each step carefully considers the conditions under which the original functions and their combinations are defined.
