00:01
Let's find the absolute minimum and absolute maximum values of the function f of t equals t plus co -tangent of t -half on the interval by fourth, seven by fourth.
00:18
So first we're going to remember the definition of co -tangent of t -half.
00:26
We know that that's one over tangent of t -half.
00:32
And because tangent of t -half is sine of t -half over cosine of t -half, we will have here t -plus cosine of t -half over sign of t -half.
00:49
It means that the function is defined where whenever sign of t -half is not zero.
01:00
But in this case t is in the interval from p -p -falfth or through 7 by 4th.
01:12
And that means that pi 4th is less than or equal to t less than or equal to 7 5th.
01:28
That means that by 8th is less than or equal to t half, less than or equal to 7 by 8th.
01:40
And we know that this value here, pi eighth is positive, of course, because pi is a positive number, 8 is a positive number.
01:54
And this is less than pi, because we have a multiple with a scalar 7 over 8, which is less than 1.
02:02
So we have this inequality that means that when t is in this interval, as is the case in this problem, the argument of the sign of t half, the argument of sign of t half in this denominator here, that argument is between zero and pi, but never is never equal to zero and never equal to pi.
02:29
So sign of that argument t half is like the graph of the sign function, but a little bit after zero and a little bit before pi.
02:47
That is here is zero.
02:49
Here is pi.
02:50
Let's.
02:50
Here is y.
02:50
Say and so the graph of the sign of t half will be something like this it never reaches zero either way near zero or nearby so we can say that sign of t half is positive for all t in the interval from pi fourth to seven five fourth so that's the first thing that is is a function is well defined, that is the co -tangent of t -half never gives an undefine or non -finite number.
03:44
But also we know that because of that, the function is continuous on the interval given here.
03:54
So f is continuous on pi fourth, 7 pi fourth, for the same by fourth, for the same reason that this expression here is continuous.
04:14
T is continuous, cosine of t half is continuous, and this expression is continuous and the quotient of the two expression, cosine of t -half, and set of t -half is continuous because the denominator, sign of t -half, never is equal to zero.
04:29
So this expression is the whole function, f of t is continuous on the clause interval by fourth, seven by fourth.
04:38
And for that reason, being continued, the function and close the interval, we know that f attains its extreme values on the given interval.
05:00
It means there are points or values in this interval for which the image of those points are equal to the extreme values of the function.
05:11
We also know that these stream values are attained at either the end points of the interval or at critical numbers of the function.
05:21
So we gotta find the critical numbers of f.
05:25
And for that, we calculate first the derivative of f, and this is equal to 1 plus the derivative of cotangent of t -half, that is negative co -sequent square of t -half times the derivative of t -half is one -half.
05:49
So we get 1 minus 1 half of kosecond square of t half.
05:59
And that's equal to 1 minus co -sigan square is 1 over sine square of the 1.
06:06
So we get 1 over 2 sine square of the half.
06:14
And we know already that this, as we said before, this expression, sine of t -half is never 0 for t.
06:23
Belong into the interval pi -fold and by -forth, that's what we see here.
06:29
And for that same reason, this derivative exists for every value of t in the interval of five -folds, seven -by -folds.
06:47
And that's simply because the denominator 2 -ssquare of t -half is never zero.
06:57
And that's because sine of t -half is not zero.
07:00
For t in this interval.
07:05
Because of t half, we studied over here.
07:10
And we saw it's never zero for t on the interval 5 fourths, 7 by fourth.
07:16
So if this derivative exists for every value of t in this interval, it means then that the only critical numbers of this function in this interval are those values of t for which derivative is zero.
07:32
So we can write that now.
07:34
So we have this and we got then to solve this equation.
07:46
That is we start with the derivative of f is equal zero and that's the same thing as using the formulae of the derivative which is this one here.
07:59
1 minus 1 over 2 sine square of t half is 0.
08:07
This is the same as 2 sine square of t half equal 1 and this is the same as sine square of t half equals 1 half and this is even equivalent to sign of t half equal more or less 1 over square root of 2 which is equal to more or less right, square root of 2 over 2.
08:45
So this is the equation we're going to solve, and we now the sign of some angle is equal to a square root of 2 over 2, the angle got to be by 4th.
08:57
But there are two angles that solve these.
09:01
First of all, let's put things in perspective.
09:07
We have two equations here, sine of t -half equal square root 2 over 2, and sign of 3 -5 equal negative, square to 2 over 2.
09:16
But we saw here, we said it just here, that sign of t over 2 is positive for t in the interval pi -forths and by fourths.
09:30
That's what we saw here up.
09:34
Because t -half for t in the interval pi -forths and by fourth, t -half is strictly positive and less than bitton pi.
09:43
So we have this situation.
09:45
The sign is always positive, never reaching the value zero.
09:53
So sign of t half is positive.
09:55
So this equation is in reality only one equation, not two, because the negative sign has no sense for t in the interval, negative bifold, and bifold.
10:09
Let's say it again here.
10:15
Since this is positive for all t in the interval, we have to solve the equation, sine of t half equal square root of 2 over 2, 4, t n pi 4th, 7 pi 4th.
10:51
So now we get to look a little bit at the graph of sine of x, that is the sine function...