High School

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Question 6: 15 Marks

The functions \( h(x) \) and \( l(x) \) are defined by:

\[ h(x) = \frac{3}{3 - 2x} \]

\[ l(x) = \frac{2x}{2x - 3} \]

Suppose the symbols \( D_h \) and \( D_l \) denote the domains of \( h \) and \( l \) respectively. Determine and simplify the equation that defines:

(6.1) The domains of \( h \) and \( l \) respectively, \( D_h \) and \( D_l \).

(6.2) The domain of \( h + l \) and give the set \( D_{h+l} \).

Answer :

To determine the domains for the functions [tex]h(x)[/tex] and [tex]l(x)[/tex], we need to find the values of [tex]x[/tex] that make the functions undefined. A function is undefined when its denominator is equal to zero.

(6.1) Finding [tex]D_h[/tex] and [tex]D_l[/tex]:


  1. Function (h(x) = \frac{3}{3 - 2x}:

    For [tex]h(x)[/tex] to be defined, the denominator [tex]3 - 2x[/tex] must not be zero.

    Solve for [tex]x:[/tex]

    [tex]3 - 2x \neq 0[/tex]

    [tex]-2x \neq -3[/tex]

    [tex]x \neq \frac{3}{2}[/tex]

    Therefore, the domain [tex]D_h[/tex] is all real numbers except [tex]x = \frac{3}{2}[/tex].


  2. Function (l(x) = \frac{2x}{2x - 3}:

    For [tex]l(x)[/tex] to be defined, the denominator [tex]2x - 3[/tex] must not be zero.

    Solve for [tex]x:[/tex]

    [tex]2x - 3 \neq 0[/tex]

    [tex]2x \neq 3[/tex]

    [tex]x \neq \frac{3}{2}[/tex]

    Therefore, the domain [tex]D_l[/tex] is all real numbers except [tex]x = \frac{3}{2}[/tex].



(6.2) Finding the domain of [tex]h + l[/tex], (D_{h+l}:

The sum [tex]h(x) + l(x)[/tex] is defined wherever both [tex]h(x)[/tex] and [tex]l(x)[/tex] are defined. Thus, the domain of [tex]h + l[/tex] is the intersection of [tex]D_h[/tex] and [tex]D_l[/tex].

Since both [tex]D_h[/tex] and [tex]D_l[/tex] exclude [tex]x = \frac{3}{2}[/tex], the domain [tex]D_{h+l}[/tex] is:


  • All real numbers except [tex]x = \frac{3}{2}[/tex].


In simple terms, the expressions are defined when [tex]x[/tex] is any real number except [tex]\frac{3}{2}[/tex].

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