[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.27,0:00:01.94,Default,,0000,0000,0000,,- [Instructor] The table\Ngives selected values
Dialogue: 0,0:00:01.94,0:00:03.98,Default,,0000,0000,0000,,of the differentiable function f.
Dialogue: 0,0:00:03.98,0:00:07.06,Default,,0000,0000,0000,,All right, can we use\Nthe mean value theorem
Dialogue: 0,0:00:07.06,0:00:10.30,Default,,0000,0000,0000,,to say that there is a value\Nc such that f prime of c
Dialogue: 0,0:00:10.30,0:00:13.03,Default,,0000,0000,0000,,is equal to five and c\Nis between four and six?
Dialogue: 0,0:00:13.03,0:00:15.30,Default,,0000,0000,0000,,If so, write a justification.
Dialogue: 0,0:00:15.30,0:00:19.00,Default,,0000,0000,0000,,Well it meets, to meet the,\Nto use the mean value theorem,
Dialogue: 0,0:00:19.00,0:00:21.25,Default,,0000,0000,0000,,you have to be differentiable\Nover the open interval
Dialogue: 0,0:00:21.25,0:00:23.63,Default,,0000,0000,0000,,and continuous over the closed interval,
Dialogue: 0,0:00:23.63,0:00:25.00,Default,,0000,0000,0000,,so it seems like we've met that.
Dialogue: 0,0:00:25.00,0:00:26.88,Default,,0000,0000,0000,,Because if you're\Ndifferentiable over an interval,
Dialogue: 0,0:00:26.88,0:00:28.67,Default,,0000,0000,0000,,you're definitely continuous\Nover that interval.
Dialogue: 0,0:00:28.67,0:00:31.82,Default,,0000,0000,0000,,It's saying that it's just\Na generally differentiable
Dialogue: 0,0:00:31.82,0:00:33.50,Default,,0000,0000,0000,,function f I guess over any interval.
Dialogue: 0,0:00:33.50,0:00:35.08,Default,,0000,0000,0000,,But the next part is to say,
Dialogue: 0,0:00:35.08,0:00:38.06,Default,,0000,0000,0000,,all right, if that condition is met,
Dialogue: 0,0:00:38.06,0:00:39.68,Default,,0000,0000,0000,,then the slope of the secant line
Dialogue: 0,0:00:39.68,0:00:43.94,Default,,0000,0000,0000,,between four comma f of\Nfour and six comma f of six,
Dialogue: 0,0:00:43.94,0:00:48.10,Default,,0000,0000,0000,,that some at least one point\Nin between four and six
Dialogue: 0,0:00:48.10,0:00:49.86,Default,,0000,0000,0000,,will have a derivative that is equal
Dialogue: 0,0:00:49.86,0:00:51.41,Default,,0000,0000,0000,,to the slope of the secant line.
Dialogue: 0,0:00:51.41,0:00:52.75,Default,,0000,0000,0000,,And so let's figure out what the slope
Dialogue: 0,0:00:52.75,0:00:55.89,Default,,0000,0000,0000,,of the secant line is\Nbetween four comma f of four
Dialogue: 0,0:00:55.89,0:00:57.29,Default,,0000,0000,0000,,and six comma f of six.
Dialogue: 0,0:00:57.29,0:00:59.09,Default,,0000,0000,0000,,And if it's equal to five,
Dialogue: 0,0:00:59.09,0:01:01.14,Default,,0000,0000,0000,,then we could use the mean value theorem.
Dialogue: 0,0:01:02.08,0:01:02.91,Default,,0000,0000,0000,,If it's not equal to five,
Dialogue: 0,0:01:02.91,0:01:04.04,Default,,0000,0000,0000,,then the mean value\Ntheorem would not apply.
Dialogue: 0,0:01:04.04,0:01:05.41,Default,,0000,0000,0000,,And so let's do that.
Dialogue: 0,0:01:05.41,0:01:08.60,Default,,0000,0000,0000,,f of six minus f of four,
Dialogue: 0,0:01:08.60,0:01:12.38,Default,,0000,0000,0000,,all of that over six\Nminus four is equal to
Dialogue: 0,0:01:12.38,0:01:13.91,Default,,0000,0000,0000,,seven minus three
Dialogue: 0,0:01:14.96,0:01:17.13,Default,,0000,0000,0000,,over two, which is equal to two.
Dialogue: 0,0:01:17.13,0:01:20.27,Default,,0000,0000,0000,,So two not equal to five.
Dialogue: 0,0:01:20.27,0:01:24.85,Default,,0000,0000,0000,,So mean value theorem doesn't apply.
Dialogue: 0,0:01:26.74,0:01:29.24,Default,,0000,0000,0000,,All right let's, I'll\Nput an exclamation mark
Dialogue: 0,0:01:29.24,0:01:31.41,Default,,0000,0000,0000,,there for emphasis.
Dialogue: 0,0:01:31.41,0:01:32.65,Default,,0000,0000,0000,,All right let's do the next part.
Dialogue: 0,0:01:32.65,0:01:34.97,Default,,0000,0000,0000,,Can we use the mean value theorem to say
Dialogue: 0,0:01:34.97,0:01:36.65,Default,,0000,0000,0000,,that the equation f prime of\Nx is equal to negative one
Dialogue: 0,0:01:36.65,0:01:37.62,Default,,0000,0000,0000,,has a solution?
Dialogue: 0,0:01:37.62,0:01:39.30,Default,,0000,0000,0000,,And now the interval's from zero to two.
Dialogue: 0,0:01:39.30,0:01:41.14,Default,,0000,0000,0000,,If so, write a justification.
Dialogue: 0,0:01:41.14,0:01:42.65,Default,,0000,0000,0000,,All right so let's see this.
Dialogue: 0,0:01:42.65,0:01:45.94,Default,,0000,0000,0000,,So if we were to take the\Nslope of the secant line,
Dialogue: 0,0:01:45.94,0:01:49.16,Default,,0000,0000,0000,,so f of two minus f of zero,
Dialogue: 0,0:01:49.16,0:01:51.07,Default,,0000,0000,0000,,all that over two minus zero.
Dialogue: 0,0:01:51.07,0:01:53.99,Default,,0000,0000,0000,,This is equal to negative two minus zero,
Dialogue: 0,0:01:54.83,0:01:56.38,Default,,0000,0000,0000,,all of that over two,
Dialogue: 0,0:01:56.38,0:01:57.77,Default,,0000,0000,0000,,which is equal to negative two over two,
Dialogue: 0,0:01:57.77,0:02:00.40,Default,,0000,0000,0000,,which is equal to negative one.
Dialogue: 0,0:02:00.40,0:02:04.25,Default,,0000,0000,0000,,And so and we also know that we meet
Dialogue: 0,0:02:04.25,0:02:06.66,Default,,0000,0000,0000,,the continuity and\Ndifferentiability conditions,
Dialogue: 0,0:02:06.66,0:02:10.28,Default,,0000,0000,0000,,and so we could say and since
Dialogue: 0,0:02:11.77,0:02:15.80,Default,,0000,0000,0000,,f is generally differentiable,
Dialogue: 0,0:02:15.80,0:02:18.48,Default,,0000,0000,0000,,generally differentiable,
Dialogue: 0,0:02:19.35,0:02:20.59,Default,,0000,0000,0000,,differentiable,
Dialogue: 0,0:02:21.83,0:02:25.47,Default,,0000,0000,0000,,it will be
Dialogue: 0,0:02:25.47,0:02:27.77,Default,,0000,0000,0000,,differentiable, differentiable,
Dialogue: 0,0:02:29.49,0:02:32.62,Default,,0000,0000,0000,,and, and continuous
Dialogue: 0,0:02:33.55,0:02:36.68,Default,,0000,0000,0000,,over the interval from zero to two,
Dialogue: 0,0:02:36.68,0:02:37.62,Default,,0000,0000,0000,,and I'll say the closed interval.
Dialogue: 0,0:02:37.62,0:02:39.59,Default,,0000,0000,0000,,You just have to be differentiable\Nover the open interval,
Dialogue: 0,0:02:39.59,0:02:41.72,Default,,0000,0000,0000,,but it's even better I guess\Nif you're differentiable
Dialogue: 0,0:02:41.72,0:02:43.51,Default,,0000,0000,0000,,over the closed interval because\Nyou have to be continuous
Dialogue: 0,0:02:43.51,0:02:44.92,Default,,0000,0000,0000,,over the closed interval.
Dialogue: 0,0:02:44.92,0:02:47.33,Default,,0000,0000,0000,,And so, and since f is\Ngenerally differentiable,
Dialogue: 0,0:02:47.33,0:02:51.20,Default,,0000,0000,0000,,it will be differentiable\Nand continuous over zero two.
Dialogue: 0,0:02:51.20,0:02:55.13,Default,,0000,0000,0000,,So the mean value theorem tells us,
Dialogue: 0,0:02:55.13,0:02:58.98,Default,,0000,0000,0000,,tells us, that there
Dialogue: 0,0:02:58.98,0:03:01.61,Default,,0000,0000,0000,,is an x
Dialogue: 0,0:03:01.61,0:03:04.96,Default,,0000,0000,0000,,in that interval from zero to two
Dialogue: 0,0:03:04.96,0:03:08.13,Default,,0000,0000,0000,,such that
Dialogue: 0,0:03:08.13,0:03:10.26,Default,,0000,0000,0000,,f prime of x
Dialogue: 0,0:03:10.26,0:03:13.73,Default,,0000,0000,0000,,is equal to that secant slope,
Dialogue: 0,0:03:13.73,0:03:15.56,Default,,0000,0000,0000,,or you could say that\Naverage rate of change,
Dialogue: 0,0:03:15.56,0:03:18.70,Default,,0000,0000,0000,,is equal to negative one.
Dialogue: 0,0:03:18.70,0:03:19.97,Default,,0000,0000,0000,,And so I could write,
Dialogue: 0,0:03:19.97,0:03:23.51,Default,,0000,0000,0000,,yes, yes, and then this\Nwould be my justification.
Dialogue: 0,0:03:23.51,0:03:25.15,Default,,0000,0000,0000,,This is the slope of the secant line
Dialogue: 0,0:03:25.15,0:03:26.43,Default,,0000,0000,0000,,or the average rate of change,
Dialogue: 0,0:03:26.43,0:03:28.49,Default,,0000,0000,0000,,and since f is generally differentiable,
Dialogue: 0,0:03:28.49,0:03:30.36,Default,,0000,0000,0000,,it will be differentiable and continuous
Dialogue: 0,0:03:30.36,0:03:31.56,Default,,0000,0000,0000,,over the closed interval.
Dialogue: 0,0:03:31.56,0:03:33.57,Default,,0000,0000,0000,,So the mean value theorem tells us
Dialogue: 0,0:03:33.57,0:03:35.93,Default,,0000,0000,0000,,that there is an x in this interval
Dialogue: 0,0:03:35.93,0:03:39.78,Default,,0000,0000,0000,,such that f prime of x\Nis equal to negative one.
Dialogue: 0,0:03:39.78,0:03:40.61,Default,,0000,0000,0000,,And we're done.