For questions 16, 17, and 18, the functions [tex]f[/tex], [tex]g[/tex], and [tex]h[/tex] are defined as follows:

[tex]\[
\begin{array}{l}
f : x \mapsto 1 - 2x \\
g: x \mapsto \frac{x^3}{10} \\
h: x \mapsto \frac{12}{x}
\end{array}
\][/tex]



Answer :

Certainly! Let's examine each of the functions [tex]\( f \)[/tex], [tex]\( g \)[/tex], and [tex]\( h \)[/tex] and calculate their values for [tex]\( x = 2 \)[/tex].

### Function [tex]\( f \)[/tex]

The function [tex]\( f \)[/tex] is defined as:
[tex]\[ f(x) = 1 - 2x \][/tex]

To find [tex]\( f(2) \)[/tex], we substitute [tex]\( x = 2 \)[/tex] into the function:
[tex]\[ f(2) = 1 - 2(2) \][/tex]
[tex]\[ f(2) = 1 - 4 \][/tex]
[tex]\[ f(2) = -3 \][/tex]

### Function [tex]\( g \)[/tex]

The function [tex]\( g \)[/tex] is defined as:
[tex]\[ g(x) = \frac{x^3}{10} \][/tex]

To find [tex]\( g(2) \)[/tex], we substitute [tex]\( x = 2 \)[/tex] into the function:
[tex]\[ g(2) = \frac{2^3}{10} \][/tex]
[tex]\[ g(2) = \frac{8}{10} \][/tex]
[tex]\[ g(2) = 0.8 \][/tex]

### Function [tex]\( h \)[/tex]

The function [tex]\( h \)[/tex] is defined as:
[tex]\[ h(x) = \frac{12}{x} \][/tex]

To find [tex]\( h(2) \)[/tex], we substitute [tex]\( x = 2 \)[/tex] into the function:
[tex]\[ h(2) = \frac{12}{2} \][/tex]
[tex]\[ h(2) = 6 \][/tex]

### Summary

After evaluating each of the functions at [tex]\( x = 2 \)[/tex], we get the following results:
[tex]\[ f(2) = -3 \][/tex]
[tex]\[ g(2) = 0.8 \][/tex]
[tex]\[ h(2) = 6 \][/tex]

So, the values of the functions [tex]\( f \)[/tex], [tex]\( g \)[/tex], and [tex]\( h \)[/tex] at [tex]\( x = 2 \)[/tex] are [tex]\( -3 \)[/tex], [tex]\( 0.8 \)[/tex], and [tex]\( 6 \)[/tex] respectively.