Suppose that the functions [tex]g[/tex] and [tex]h[/tex] are defined for all real numbers [tex]x[/tex] as follows.

[tex]\[
\begin{array}{l}
g(x) = 2x - 3 \\
h(x) = 6x
\end{array}
\][/tex]

Write the expressions for [tex](h - g)(x)[/tex] and [tex](h \cdot g)(x)[/tex] and evaluate [tex](h + g)(1)[/tex].

[tex]\[
\begin{array}{l}
(h - g)(x) = \square \\
(h \cdot g)(x) = \square \\
(h + g)(1) = \square
\end{array}
\][/tex]



Answer :

Let's start by defining the functions [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex]:

[tex]\[ g(x) = 2x - 3 \][/tex]
[tex]\[ h(x) = 6x \][/tex]

Now, we need to find the expressions for [tex]\( (h - g)(x) \)[/tex] and [tex]\( (h \cdot g)(x) \)[/tex], and evaluate [tex]\( (h + g)(1) \)[/tex].

1. Expression for [tex]\( (h - g)(x) \)[/tex]:

[tex]\[ (h - g)(x) = h(x) - g(x) \][/tex]

Substitute the given functions [tex]\( h(x) \)[/tex] and [tex]\( g(x) \)[/tex]:

[tex]\[ (h - g)(x) = 6x - (2x - 3) \][/tex]

Distribute the negative sign inside the parentheses:

[tex]\[ (h - g)(x) = 6x - 2x + 3 \][/tex]

Simplify the expression:

[tex]\[ (h - g)(x) = 4x + 3 \][/tex]

So, we have:

[tex]\[ (h - g)(x) = 4x + 3 \][/tex]

2. Expression for [tex]\( (h \cdot g)(x) \)[/tex]:

[tex]\[ (h \cdot g)(x) = h(x) \cdot g(x) \][/tex]

Substitute the given functions [tex]\( h(x) \)[/tex] and [tex]\( g(x) \)[/tex]:

[tex]\[ (h \cdot g)(x) = 6x \cdot (2x - 3) \][/tex]

Distribute [tex]\( 6x \)[/tex]:

[tex]\[ (h \cdot g)(x) = 12x^2 - 18x \][/tex]

So, we have:

[tex]\[ (h \cdot g)(x) = 12x^2 - 18x \][/tex]

3. Evaluate [tex]\( (h + g)(1) \)[/tex]:

[tex]\[ (h + g)(1) = h(1) + g(1) \][/tex]

First, evaluate [tex]\( h(1) \)[/tex]:

[tex]\[ h(1) = 6 \cdot 1 = 6 \][/tex]

Next, evaluate [tex]\( g(1) \)[/tex]:

[tex]\[ g(1) = 2 \cdot 1 - 3 = 2 - 3 = -1 \][/tex]

Now sum these values:

[tex]\[ (h + g)(1) = 6 + (-1) = 5 \][/tex]

So, we have:

[tex]\[ (h + g)(1) = 5 \][/tex]

To summarize, the final answers are:

[tex]\[ \begin{array}{l} (h - g)(x) = 4x + 3 \\ (h \cdot g)(x) = 12x^2 - 18x \\ (h + g)(1) = 5 \end{array} \][/tex]

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