Problem 7
x is the number of tickets and f(x) is the cost of those x tickets. So if you wanted say 2 tickets, then you would plug x = 2 into the f(x) function to find the cost it would take to get those 2 tickets.
The inverse function reverses all this. If we have the cost in mind, then we can use the inverse to determine how many tickets cost that dollar amount. Since everything swaps, this means x is now the cost in dollars and f^(-1)(x) is the number of tickets. Problem 8 below explains this in more detail while also going over an example.
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Problem 8
You should find back in problem 6 that the inverse function is
f^(-1)(x) = (x-12)/44
Let's plug in x = 320
f^(-1)(x) = (x-12)/44
f^(-1)(320) = (320-12)/44 ... replace x with 320
f^(-1)(320) = 308/44
f^(-1)(320) = 7
So x = 320 leads to f^(-1)(x) = 7
Meaning that if you have a budget of $320, then you can buy at most 7 tickets. Or if you want to spend exactly $320, then you need to buy 7 tickets.
Back in problem 5, the f(x) function you should have gotten was f(x) = 44x+12
Let's plug in x = 7 and see what happens
f(x) = 44x+12
f(7) = 44*7+12
f(7) = 308+12
f(7) = 320
So this shows that 7 tickets cost $320, which is the reverse of what we got earlier. Which is expected because the inverse swaps everything.
So in summary, f^(-1)(320) = 7 tells the reader "if you want to spend $320, then you need to get 7 tickets". In a sense, it is the same as solving 44x+12 = 320 for x to get x = 7.
