0:00
All right.
00:01
So we're given a function m of r that tells us the monthly payment associated with a given interest rate.
00:12
And then we know that there's rates offered in the range from 0 .04 to 0 .05.
00:21
In fact, we can make those inclusive with square brackets.
00:26
And we're interested in finding a monthly payment of $1 ,300.
00:32
So we want to know if there's a rate in that, in the given range where the monthly payment will be $1 ,300.
00:40
So the first thing we're going to do is try to show that there is, in fact, a solution.
00:49
And in order to do that, the goal is to, of course, use the intermediate value theorem.
00:53
So in order to satisfy the intermediate value theorem, we need to know that that m of r is continuous on this range from 0 .04 to 0 .05.
01:07
So to do that, since it's more or less a giant fraction, we just need to check that the denominator is never zero in that range.
01:21
So we'll write out the denominator here, 1 minus 1 plus r over 12 to the negative 360.
01:32
We're going to set that equal to zero.
01:34
And so we want to show that this equation doesn't have any solutions.
01:39
So we'll just move the large term to the other side, adding it to both sides.
01:45
One plus r over 12 to the negative 360.
01:51
And then what we can do is raise both sides of the equation to the negative 1 over 360 of power to get rid of this exponent.
02:04
Since the left -hand side is just one, this won't be too much trouble.
02:12
So on the right -hand side, we can cross those out.
02:14
But remember that we're taking an even root here, 360.
02:20
So that's going to give us plus or minus one as opposed to just one as the root.
02:29
And then the negative exponent just means that we're basically taking one over the result, which in this case is still just one or negative one.
02:39
So to write that out, we're going to have either plus or minus one equal to what's left on the right hand side.
02:52
So we can split this up.
02:59
So either positive 1 equals 1 plus r over 12, or negative 1 equals 1 plus r over 12.
03:12
If either of these has a solution, then there's potentially a continuity problem in our function.
03:18
So starting with the first one, we subtract one from both sides.
03:22
0 equals r over 12.
03:26
Since r is between 0 .04 and 05, this obviously has no solutions.
03:31
R is never zero in that range.
03:33
And then for the second equation, we can add one to both sides.
03:37
We get 0 equals 2 plus r over 12...