00:01
We will find the absolute minimum and absolute maximum values as a function f of t equals of t plus sine of 2 times t on the closed interval from 0 to pi half.
00:14
We know this function is continuous in this close interval, and for that reason we know f attains its extreme values at points on that interval.
00:26
In fact, we know those points where f attain its extreme values may be either the end point, of the interval or critical numbers of the function.
00:35
For that reason, we need to calculate the critical numbers of f.
00:40
For that, we calculate first the first order derivative of f, and that is negative 2 sine of t plus 2 cosine of 2t.
00:58
And now this derivative is defined everywhere in the real number, so f derivative of t exists for every t in the interval 0 by half.
01:16
And for that reason, we can say that the only critical numbers of f are those values of t for which the first derivative is equal to zero.
01:31
For that reason, we get to solve this equation, first derivative of f equal to zero.
01:38
And this is equivalent to the equation negative 2 sign of t plus two cosine of two t equals zero which is the same as cosine of two t equal or in fact we can say instead of that all time of two t minus end of t equal to zero and now we can use the formula of the cosine of the double of an angle which we know is equal to cosine square of t minus sine square of t and that minus sine of t equals zero.
02:37
And now cosine of square of t is 1 minus sine square of t and that minus sine square of t minus sine of t equals zero.
02:53
And then this is equivalent to the equation 1 minus 2 sine square of t minus sine of t equals zero, which can be rewritten as 2 sine square of t plus sine of t minus 1 equals 0.
03:23
And now we can make a change of variable here.
03:29
If we name alpha equals sine of t, or maybe let's not use alphabet, another letter, satin letter like m, then we get to m square plus m minus 1 equals 0.
03:55
And we can solve this equation using the formula we know for the solutions of this equation, which is equal to negative 1 more or less square root of 1 square, which is 1 minus 4 times 2 times negative 1 over 2 times 2.
04:16
That gives us negative 1 more or less square root of 1 plus 8 over 4 that give us negative 1 more or less square root of 9 over 4 that is negative 1 more or less 3 over 4 so we get m1 is negative 1 plus 3 over 4 which is 3 minus 1 is 2 over 4 is 1 1 half and m2 is negative 1 minus 4 over 4 and that is minus 3 i meant sorry here minus 3 and that is negative 4 over 4 is negative 1 so remember m is sign of t so we get sign of t equal 1 1 1 half or sine of t equal negative 1 for t in the interval we remember the domain is 0 by half.
05:48
Okay, so we got to remember the assign function a little bit here.
05:56
Over the interval 0 by half, we have sign of 0 is 0, so again, it increases.
06:05
Let me throw this smaller here.
06:12
So let's say get this, at pi half, we know sign is equal to 1, and at 0 is 0.
06:22
So here we can say the value is 1 here.
06:29
So this is sign of t for this interval.
06:35
And at that interval, this solution has no equation.
06:40
That's no solution.
06:41
This equation has no solution because we have a sign always positive or zero.
06:46
So this cannot be.
06:52
And so sign of t equal one half is the angle for which we have this.
06:59
Value one half here, and we know sign of 5x is one half.
07:15
And that reason we can say that t is 5 .6.
07:22
And it's the only solution, in fact, because graphically we can see that there are no the points in the interval 0 to pi half, where a sign of t is 1 half.
07:35
In fact, the function is increasing, so it's one to one over there.
07:41
So there is only one value of t between 0 and pi half, for which the image is equal to 1 .5.
07:49
And that number is pi 6 belonging to this interval 0 .5.
07:55
So we have this a critical number, and that's the only critical number, as we said here above...