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A tuning fork vibrates with a frequency of 440 hertz (cycles/second). When the tuning fork is struck, it produces a change in the normal air pressure in the room.

Function [tex]$p$[/tex] represents this situation, where [tex]$p(t)$[/tex] is the change in pressure, in pascals, relative to the normal air pressure in the room as a function of time, [tex]$t$[/tex], in seconds, after the tuning fork is struck.
[tex]$
p(t) = 5 \sin (880 \pi t)
$[/tex]

What are the domain and the range within the context of this situation?

The domain of the function is [tex]$t \geq \ \square$[/tex]

The range of the function is [tex]$\square \leq p(t) \leq \square$[/tex].

Sagot :

GinnyAnsweridn
To determine the domain and range of the function [tex]\( p(t) = 5 \sin(880 \pi t) \)[/tex] in the context of the given situation, follow these steps:

1. Determine the domain:
- The function [tex]\( p(t) \)[/tex] represents the change in air pressure over time [tex]\( t \)[/tex].
- Time [tex]\( t \)[/tex] cannot be negative, as it represents the duration after the tuning fork is struck.
- Therefore, the domain is [tex]\( t \geq 0 \)[/tex].

2. Determine the range:
- The sine function, [tex]\( \sin(\theta) \)[/tex], has a range between -1 and 1.
- Given function is [tex]\( 5 \sin(880 \pi t) \)[/tex], the sine function is scaled by a factor of 5.
- Hence, the minimum value of [tex]\( 5 \sin(880 \pi t) \)[/tex] is [tex]\( 5 \times (-1) = -5 \)[/tex].
- The maximum value of [tex]\( 5 \sin(880 \pi t) \)[/tex] is [tex]\( 5 \times 1 = 5 \)[/tex].

Thus, we can fill in the blanks with these values:
- The domain of the function is [tex]\( t \geq 0 \)[/tex].
- The range of the function is [tex]\( -5 \leq p(t) \leq 5 \)[/tex].

So, the correct answers for the boxes are:

The domain of the function is [tex]\( t \geq 0 \)[/tex].

The range of the function is [tex]\( -5 \leq p(t) \leq 5 \)[/tex].
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