Answer:
The domain of the function b(s) is
![s\in[0,150]](https://img.qammunity.org/2021/formulas/mathematics/college/40egdq3x0pb7q8u49s9ruh189dc6yebyzy.png)
Explanation:
Given that, a total of 560 books is added to the book shelf each morning.
be the number of the students who visit the library on a particular day and takes exactly 3 books from this shelf.
So, the number of books they take from the shelf is 3s.
The number of remaining books the shelf
.
As , given that
be the number of books left on the shelf at the end of that day, so the required function,
, is
![b=560-3s\;\cdots(i)](https://img.qammunity.org/2021/formulas/mathematics/college/y09tdl86zf70z72xy3jmpat1zsqqqmmuzs.png)
As there are 150 students in the school. So, if no one will go to the library, than
which is the minimum value, and if all goes to the library , than
which is the maximum value of
.
So, the possible value of s is:
![0\leq s\leq150\;\cdots(ii)](https://img.qammunity.org/2021/formulas/mathematics/college/endb0gws9oh1ycga6ur6pmac4dhs1m9u8y.png)
Now, as there is no book left or there are some books left the negative value of
is not possible. So,
![b\geq0](https://img.qammunity.org/2021/formulas/mathematics/college/qw4mbovvubao0iwo9zqux3mfeoivfhrb6c.png)
[fron equation (i)]
![\Rightarrow s\leq 560/3](https://img.qammunity.org/2021/formulas/mathematics/college/ynm93gckv9e0hvz7xogsud4rojmr1j6zi1.png)
![\Rightarrow s\leq 186(2)/(3)](https://img.qammunity.org/2021/formulas/mathematics/college/a86q9rvxxg9zybqtln1729ay7252sv6mlj.png)
as s id the number of students which cant be a fractional value, so the possible nearest value is,
![s\leq 186\;\cdots(iii)](https://img.qammunity.org/2021/formulas/mathematics/college/kibfl9lalunffrlco3j3ww47400o4s5oj5.png)
From the equations (ii) and (iii), the domain of the function
is
![s\in[0,150]](https://img.qammunity.org/2021/formulas/mathematics/college/40egdq3x0pb7q8u49s9ruh189dc6yebyzy.png)