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37 votes
37 votes
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.

Your pet groomer charges according to how much your dog weighs. For a dog that weighs-example-1
Your pet groomer charges according to how much your dog weighs. For a dog that weighs-example-1
Your pet groomer charges according to how much your dog weighs. For a dog that weighs-example-2
User FarukT
by
3.2k points

1 Answer

17 votes
17 votes

Step-by-step explanation

Given: The pet groomer charges according to the dog weights as follows:


\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}

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

This is achieved thus:

==> For the first function


\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}

==> For the second function


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

==> For the third function


\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}

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.

User Rob Bonner
by
2.8k points
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