[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.26,0:00:03.65,Default,,0000,0000,0000,,- [Voiceover] The figure above\Nshows the graph of f prime,
Dialogue: 0,0:00:03.65,0:00:06.61,Default,,0000,0000,0000,,the derivative of a\Ntwice-differentiable function f,
Dialogue: 0,0:00:06.61,0:00:08.05,Default,,0000,0000,0000,,on the interval, that's a closed interval,
Dialogue: 0,0:00:08.05,0:00:09.58,Default,,0000,0000,0000,,from negative three to four.
Dialogue: 0,0:00:09.58,0:00:12.26,Default,,0000,0000,0000,,The graph of f prime\Nhas horizontal tangents
Dialogue: 0,0:00:12.26,0:00:14.53,Default,,0000,0000,0000,,at x equals negative one, x equals one,
Dialogue: 0,0:00:14.53,0:00:15.86,Default,,0000,0000,0000,,and x equals three.
Dialogue: 0,0:00:15.86,0:00:19.19,Default,,0000,0000,0000,,So you have a horizontal\Ntangent right over,
Dialogue: 0,0:00:19.19,0:00:21.92,Default,,0000,0000,0000,,a horizontal tangent right over there.
Dialogue: 0,0:00:23.04,0:00:24.52,Default,,0000,0000,0000,,And let me draw that a little bit neater,
Dialogue: 0,0:00:24.52,0:00:25.46,Default,,0000,0000,0000,,right over there,
Dialogue: 0,0:00:25.46,0:00:27.32,Default,,0000,0000,0000,,a horizontal tangent right over there,
Dialogue: 0,0:00:27.32,0:00:29.36,Default,,0000,0000,0000,,and a horizontal tangent right over there.
Dialogue: 0,0:00:30.98,0:00:31.56,Default,,0000,0000,0000,,Alright.
Dialogue: 0,0:00:31.56,0:00:33.34,Default,,0000,0000,0000,,The areas of the regions bounded
Dialogue: 0,0:00:33.34,0:00:36.48,Default,,0000,0000,0000,,by the x-axis and the graph of f prime
Dialogue: 0,0:00:36.48,0:00:38.73,Default,,0000,0000,0000,,on the intervals negative two to one,
Dialogue: 0,0:00:38.73,0:00:40.35,Default,,0000,0000,0000,,closed intervals from negative two to one,
Dialogue: 0,0:00:40.35,0:00:42.69,Default,,0000,0000,0000,,so this region right over here,
Dialogue: 0,0:00:42.69,0:00:45.14,Default,,0000,0000,0000,,and the region from one to four,
Dialogue: 0,0:00:45.14,0:00:46.59,Default,,0000,0000,0000,,so this region right over there,
Dialogue: 0,0:00:46.59,0:00:48.49,Default,,0000,0000,0000,,they tell us have,
Dialogue: 0,0:00:48.49,0:00:50.91,Default,,0000,0000,0000,,have the areas are 9 and 12, respectively.
Dialogue: 0,0:00:50.91,0:00:54.20,Default,,0000,0000,0000,,So that area's 9 and that area is 12.
Dialogue: 0,0:00:54.20,0:00:55.73,Default,,0000,0000,0000,,So Part A,
Dialogue: 0,0:00:55.73,0:00:59.96,Default,,0000,0000,0000,,Find all x coordinates at\Nwhich f has a relative maximum.
Dialogue: 0,0:00:59.96,0:01:03.47,Default,,0000,0000,0000,,Give a reason for your answer.
Dialogue: 0,0:01:03.47,0:01:05.53,Default,,0000,0000,0000,,All x-coordinates at which f
Dialogue: 0,0:01:05.53,0:01:06.96,Default,,0000,0000,0000,,has a relative maximum.
Dialogue: 0,0:01:06.96,0:01:07.63,Default,,0000,0000,0000,,So you might say,
Dialogue: 0,0:01:07.63,0:01:09.20,Default,,0000,0000,0000,,"Oh wait, wait this looks\Nlike a relative maximum
Dialogue: 0,0:01:09.20,0:01:10.51,Default,,0000,0000,0000,,over here, but this isn't f.
Dialogue: 0,0:01:10.51,0:01:12.41,Default,,0000,0000,0000,,This is the graph of f prime."
Dialogue: 0,0:01:12.41,0:01:13.79,Default,,0000,0000,0000,,So let's think about when we used to
Dialogue: 0,0:01:13.79,0:01:15.49,Default,,0000,0000,0000,,we don't have a graph of f in front of us.
Dialogue: 0,0:01:15.49,0:01:17.43,Default,,0000,0000,0000,,So let's think about\Nwhat it needs to be true
Dialogue: 0,0:01:17.43,0:01:20.94,Default,,0000,0000,0000,,for f to have a relative\Nmaximum at a point.
Dialogue: 0,0:01:20.94,0:01:23.21,Default,,0000,0000,0000,,We are probably familiar with
Dialogue: 0,0:01:23.21,0:01:25.42,Default,,0000,0000,0000,,what relative maximum will look like.
Dialogue: 0,0:01:25.42,0:01:28.21,Default,,0000,0000,0000,,It'd look like a little lump, like that.
Dialogue: 0,0:01:28.21,0:01:30.07,Default,,0000,0000,0000,,It could also actually look like that
Dialogue: 0,0:01:30.07,0:01:32.17,Default,,0000,0000,0000,,but since this is differentiable function
Dialogue: 0,0:01:32.17,0:01:33.67,Default,,0000,0000,0000,,over the interval, we're\Nprobably not dealing
Dialogue: 0,0:01:33.67,0:01:36.21,Default,,0000,0000,0000,,with a relative maximum\Nthat looks like that.
Dialogue: 0,0:01:37.57,0:01:41.80,Default,,0000,0000,0000,,And so what do we know about\Na relative maximum point?
Dialogue: 0,0:01:41.80,0:01:44.75,Default,,0000,0000,0000,,So let's say that's our relative maximum.
Dialogue: 0,0:01:44.75,0:01:46.88,Default,,0000,0000,0000,,As we approach our relative maximum
Dialogue: 0,0:01:46.88,0:01:50.20,Default,,0000,0000,0000,,from values, as we have x\Nvalues that are approaching
Dialogue: 0,0:01:50.20,0:01:54.04,Default,,0000,0000,0000,,the x value of our relative maximum point,
Dialogue: 0,0:01:54.04,0:01:56.66,Default,,0000,0000,0000,,as we approach it from\Nvalues below that x value,
Dialogue: 0,0:01:56.66,0:01:59.70,Default,,0000,0000,0000,,we see that we have a positive slope.
Dialogue: 0,0:01:59.70,0:02:02.84,Default,,0000,0000,0000,,Our function needs to be increasing.
Dialogue: 0,0:02:02.84,0:02:04.75,Default,,0000,0000,0000,,So over here.
Dialogue: 0,0:02:05.78,0:02:08.69,Default,,0000,0000,0000,,Over here we see f is increasing,
Dialogue: 0,0:02:08.69,0:02:11.83,Default,,0000,0000,0000,,going into the relative maximum point,
Dialogue: 0,0:02:11.83,0:02:12.73,Default,,0000,0000,0000,,f is increasing,
Dialogue: 0,0:02:12.73,0:02:15.23,Default,,0000,0000,0000,,which means that the derivative of f
Dialogue: 0,0:02:15.23,0:02:17.31,Default,,0000,0000,0000,,the derivative of f must\Nbe greater than zero.
Dialogue: 0,0:02:18.34,0:02:20.87,Default,,0000,0000,0000,,And then after we past that maximum point,
Dialogue: 0,0:02:20.87,0:02:22.49,Default,,0000,0000,0000,,after we past that maximum point,
Dialogue: 0,0:02:22.49,0:02:26.25,Default,,0000,0000,0000,,we see that our function\Nneeds to be decreasing.
Dialogue: 0,0:02:26.25,0:02:27.57,Default,,0000,0000,0000,,Do this in another color.
Dialogue: 0,0:02:27.57,0:02:29.71,Default,,0000,0000,0000,,We see that our funciton is decreasing
Dialogue: 0,0:02:29.71,0:02:33.09,Default,,0000,0000,0000,,right over here, so f decreasing,
Dialogue: 0,0:02:34.10,0:02:36.76,Default,,0000,0000,0000,,decreasing, which means
Dialogue: 0,0:02:36.76,0:02:39.68,Default,,0000,0000,0000,,that f prime of x needs\Nto be less than zero.
Dialogue: 0,0:02:40.52,0:02:43.74,Default,,0000,0000,0000,,So our relative maximum point
Dialogue: 0,0:02:43.74,0:02:46.25,Default,,0000,0000,0000,,should happen at an x value.
Dialogue: 0,0:02:46.25,0:02:47.31,Default,,0000,0000,0000,,It should happen at an x value
Dialogue: 0,0:02:47.31,0:02:50.28,Default,,0000,0000,0000,,where our first derivative transitions
Dialogue: 0,0:02:50.28,0:02:52.48,Default,,0000,0000,0000,,from being greater than zero
Dialogue: 0,0:02:52.48,0:02:55.24,Default,,0000,0000,0000,,to being less than zero.
Dialogue: 0,0:02:55.24,0:02:56.44,Default,,0000,0000,0000,,So what x value,
Dialogue: 0,0:02:56.44,0:02:57.26,Default,,0000,0000,0000,,so let me say this,
Dialogue: 0,0:02:57.26,0:03:00.74,Default,,0000,0000,0000,,so we have
Dialogue: 0,0:03:00.74,0:03:05.41,Default,,0000,0000,0000,,f has relative, let me\Njust write it shorthand,
Dialogue: 0,0:03:05.41,0:03:10.35,Default,,0000,0000,0000,,relative maximum at x values
Dialogue: 0,0:03:11.50,0:03:13.53,Default,,0000,0000,0000,,where
Dialogue: 0,0:03:13.53,0:03:16.56,Default,,0000,0000,0000,,f prime transitions,
Dialogue: 0,0:03:16.56,0:03:17.63,Default,,0000,0000,0000,,transitions,
Dialogue: 0,0:03:19.02,0:03:21.70,Default,,0000,0000,0000,,transitions from
Dialogue: 0,0:03:21.70,0:03:26.27,Default,,0000,0000,0000,,from positive, positive,
Dialogue: 0,0:03:27.87,0:03:32.71,Default,,0000,0000,0000,,to negative, to, let me write\Nthis a little bit neater,
Dialogue: 0,0:03:32.71,0:03:34.49,Default,,0000,0000,0000,,to negative.
Dialogue: 0,0:03:35.65,0:03:37.71,Default,,0000,0000,0000,,to negative.
Dialogue: 0,0:03:37.71,0:03:39.92,Default,,0000,0000,0000,,And where do we see f prime transitioning
Dialogue: 0,0:03:39.92,0:03:41.90,Default,,0000,0000,0000,,from positive to negative?
Dialogue: 0,0:03:41.90,0:03:43.93,Default,,0000,0000,0000,,Well over here we only\Nsee that happening once.
Dialogue: 0,0:03:43.93,0:03:47.14,Default,,0000,0000,0000,,We see right here f prime is\Npositive, positive, positive,
Dialogue: 0,0:03:47.14,0:03:49.08,Default,,0000,0000,0000,,and then it goes negative,\Nnegative, negative.
Dialogue: 0,0:03:49.08,0:03:50.80,Default,,0000,0000,0000,,So we see,
Dialogue: 0,0:03:50.80,0:03:54.14,Default,,0000,0000,0000,,we see f prime is positive over here,
Dialogue: 0,0:03:55.27,0:03:58.02,Default,,0000,0000,0000,,And then, right when we\Nhit x equals negative two,
Dialogue: 0,0:03:58.02,0:04:00.70,Default,,0000,0000,0000,,f prime becomes negative.
Dialogue: 0,0:04:00.70,0:04:03.85,Default,,0000,0000,0000,,F prime becomes negative.
Dialogue: 0,0:04:03.85,0:04:05.71,Default,,0000,0000,0000,,So we know that the function itself,
Dialogue: 0,0:04:05.71,0:04:06.43,Default,,0000,0000,0000,,not f prime,
Dialogue: 0,0:04:06.43,0:04:07.88,Default,,0000,0000,0000,,f must be increasing here
Dialogue: 0,0:04:07.88,0:04:10.77,Default,,0000,0000,0000,,because f prime is positive,
Dialogue: 0,0:04:10.77,0:04:13.37,Default,,0000,0000,0000,,and then our function at f
Dialogue: 0,0:04:13.37,0:04:14.52,Default,,0000,0000,0000,,is decreasing here
Dialogue: 0,0:04:14.52,0:04:17.08,Default,,0000,0000,0000,,because f prime is negative.
Dialogue: 0,0:04:18.05,0:04:20.97,Default,,0000,0000,0000,,And so this happens at x equals two.
Dialogue: 0,0:04:20.97,0:04:22.77,Default,,0000,0000,0000,,So let me write that down.
Dialogue: 0,0:04:22.77,0:04:24.87,Default,,0000,0000,0000,,This happens at x equals two.
Dialogue: 0,0:04:24.87,0:04:27.55,Default,,0000,0000,0000,,This happens,
Dialogue: 0,0:04:27.55,0:04:31.97,Default,,0000,0000,0000,,happens at x equals two.
Dialogue: 0,0:04:31.97,0:04:33.30,Default,,0000,0000,0000,,And we're done.