To determine the domain and range of the function \( P(A) \), we start with the relationship between power and current as described in the problem. The function given is:
\[ P(A) = 110 \times A \]
where \( A \) represents the current in amps, and \( P(A) \) represents the power in joules.
Part 1: Domain
The fuse trips at 20 amps, which means the motor cannot operate at or above this current. Therefore, the domain of the function is restricted to values of \( A \) that are between 0 amps and just below 20 amps. In interval notation, this is represented as:
\[ \text{Domain: } [0, 20) \]
Part 2: Range
To find the range of \( P(A) \), we need to determine the output values as \( A \) varies within its domain. When \( A = 0 \):
\[ P(0) = 110 \times 0 = 0 \text{ joules} \]
When \( A \) approaches 20 (but does not reach it):
\[ P(20) = 110 \times 20 = 2200 \text{ joules} \]
Thus, as \( A \) increases from 0 to just below 20, \( P(A) \) increases from 0 joules to just below 2200 joules. Therefore, the range of the function in interval notation is:
\[ \text{Range: } [0, 2200) \]
Final Response
- Domain: \([0, 20)\)
- Range: \([0, 2200)\)