Respuesta :

If [tex]L(d)[/tex] is the water level after [tex]d[/tex] days, then at the start [tex](d=0)[/tex] we have [tex]L(0)=21[/tex]. After the first day [tex](d=1)[/tex], the water level falls by 1.5, so that [tex]L(1)=21-1.5=19.5[/tex]. After the second day [tex](d=2)[/tex], the water level falls by a total of 2(1.5) = 3, so that [tex]L(2)=21-2(1.5)=18[/tex]. And so on.

The idea is that [tex]L(d)[/tex] has the closed form [tex]L(d)=21-1.5d[/tex] (to answer part b).

[tex]d[/tex] could be any non-negative number of days, which means we can pick [tex]d[/tex] from the set of non-negative real numbers. However, after a certain point, our function [tex]L(d)[/tex] will start returning negative values, which would translate to the water level falling below the riverbed. So for the purposes of this problem, we should first find out when [tex]L(d)[/tex] reaches 0:

[tex]L(d)=21-1.5d=0\implies d=14[/tex]

That is, the river dries up completely after 14 days. So we can say the domain of [tex]L(d)[/tex] is [tex]0\le d\le14[/tex].

The range is the set of values that [tex]L(d)[/tex] can take on for the values of [tex]d[/tex] in the domain. In this case, we start at 21 and steadily fall to 0, so the range is [tex]0\le L(d)\le21[/tex].

After 9 days, the water level falls to

[tex]L(9)=21-9(1.5)=7.5[/tex]

so in total, the water level would have subsided by a depth of 21 - 7.5 = 13.5 ft.

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