Your pet groomer charges according to how much your dog weighs. For a dog that weighs 20 pounds or less, the groomer charges $25. For a dog that weighs more than 20 pounds and less than 50 pounds, the groomer charges $45. For any dog that weighs 50 pounds or more, the groomer charges $45 plus an additional $1 for each pound over 50. Using this situation, match each piece of the piecewise function with the corresponding domain restriction.

[tex]\begin{gathered} If\text{ }weight\leqslant20,then\text{ }charge=\text{ \$}25 \\ \\ If\text{ }weight\text{ }is\text{ }between\text{ }20\text{ }and\text{ }50,then\text{ }charge=\text{ \$}45 \\ \\ If\text{ }weight\geqslant50,then\text{ }charge=\text{ \$}45+\text{ \$}1\text{ }for\text{ }each\text{ }pound\text{ }over\text{ }50 \end{gathered}[/tex]

Required: To match each piece of the piecewise function with the corresponding restriction.

This is achieved thus:

==> For the first function

[tex]\begin{gathered} f(x)=25 \\ \\ \text{ This function satisfies the condition for x is greater than zero but less than or equal to 20} \end{gathered}[/tex]

==> For the second function

[tex]\begin{gathered} f(x)=45 \\ \\ \text{ This function satisfies the condition for x is greater than 20 but less than 50.} \end{gathered}[/tex]

==> For the third function

[tex]\begin{gathered} f(x)=45+1(x-50) \\ \\ \text{ This function satisfies the condition for x is greater than or equal to 50} \end{gathered}[/tex]

Hence, the answers are:

f(x) = 25 ==> 0 less than x less than or equal to 20

f(x) = 45 ==> 20 less than x less than 50

f(x) = 45 + 1(x - 50) ==> x greater than or equal to 50.