Tutor.com The Derivative Session

Oct. 29, 2012

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Session Transcript - Math - Calculus, 10/25/2012 9:45PM - Tutor.comSession Date: 10/25/2012 9:45PM
Length: 71.4 minute(s)
Subject: Math - Calculus


System Message
[00:00:00] *** Please note: All sessions are recorded for quality control. ***

Guest (Customer)
[00:00:00] Determine the values of the number a for which the function f has no critical number: f(x) = (a² + a - 6)cos2x + (a - 2)x + cos1
[00:00:09] hello

Eduardo d (Tutor)
[00:00:22] Welcome to Tutor.com ! Let's get started.

Guest (Customer)
[00:00:40] I finished this problem, but there weren't much work to show
[00:01:19] I asked my teacher, and he said I should do a sign diagram to answer it, but I don't know how to incorporate that into the answer

Eduardo d (Tutor)
[00:02:40] All right, can I see the wok that you did so fa, so we can figure out how to incorporate the signs value?

Guest (Customer)
[00:02:46] okay

Eduardo d (Tutor)
[00:02:47] *far

Guest (Customer)
[00:02:57] i'll just write it down
[00:03:45] I fouond the derivative in terms of x, which is: f'(x) a²+a-6)(-sin2x)(2) + (a-2)
[00:04:05] set it to 0 and got
[00:04:53] i isolate the sin
[00:05:53] since the range of sine is from -1 to 1
[00:06:18] and I'm looking for where it has no critical numbers
[00:06:48] i set the left side to less than 1 and greater than -1 so it'd give me solutions that's not included the range
[00:07:31] solved for x and got (-3.5, - 2.5)

Eduardo d (Tutor)
[00:08:51] Hmm, shouldn't it be the other way around? If you're looking for those not included in the range, then it would be where it is less than -1 and greate than 1, since the range is between -1 to 1

Guest (Customer)
[00:09:14] ooh yes
[00:09:29] but
[00:09:58] nvm yes
[00:10:02] but
[00:10:07] that's all the work i have

Eduardo d (Tutor)
[00:11:27] To be honest, that's actually all the work you need. I don't even see any need to use a signs diagram at all

Guest (Customer)
[00:11:51] i have a new problem now
[00:12:12] in class, he said that the answer is (-3.5, -2.5)
[00:12:31] my signs were messed up, so now my answers are
[00:12:44] x>-2.5, and x<-3.5

Eduardo d (Tutor)
[00:14:39] Ah...I think I now follow what you teacher meant by the "signs table". he must be refeing to the signs table needed for solving inequalities
[00:14:49] *refering

Guest (Customer)
[00:15:00] how?

Eduardo d (Tutor)
[00:15:59] So before that: How did you solve for the inequalities? Did you happen to multiply 2(a+3) to both sides?

Guest (Customer)
[00:16:30] no
[00:16:53] I multiplied 2 to both sides, then got a+3=1/2
[00:17:00] then solved for a
[00:17:16] i guess i did
[00:17:30] your method makes sense
[00:17:31] lol

Eduardo d (Tutor)
[00:18:45] All right, so actually: in inequalities, if there's a variable in the denominator, you can't multiply by that vaiable, nor take the reciprocal

Guest (Customer)
[00:19:02] got it

Eduardo d (Tutor)
[00:19:46] That's because, the variable can either be positive or negative, which in that case, the switching of signs necomes ambiguous
[00:19:49] *because
[00:20:14] So this is where the signs table comes in:

Guest (Customer)
[00:20:19] ok

Eduardo d (Tutor)
[00:21:49] Oh, before the signs table, sorry to forget, we have to rewrite the inequalities first to have 0
[00:22:53] So combining this...

Guest (Customer)
[00:23:49] how did you get the denominator?
[00:24:04] oh
[00:24:05] nvm

Eduardo d (Tutor)
[00:24:35] All right, is everything clear up to this point so far?

Guest (Customer)
[00:24:41] yes

Eduardo d (Tutor)
[00:25:58] So for the first inequality, we end up with 2a+7/2(a+3) < 0

Guest (Customer)
[00:26:04] yes

Eduardo d (Tutor)
[00:27:06] Now, we'll need to get the zeroes of each factor, and place them on a number line. That is, we need the zeoes of (2a+7) and (a+3)

Guest (Customer)
[00:27:45] ohh
[00:27:52] makes sense

Eduardo d (Tutor)
[00:28:43] So from 2a+7 we get -3.5, and from a+3 we get -3

Guest (Customer)
[00:29:12] but the answer isn't -3 though, it's -2.5

Eduardo d (Tutor)
[00:30:10] We'll get to that soon enough, we're still working on the first inequality, so we're not even yet halfway through

Guest (Customer)
[00:30:26] okay

Eduardo d (Tutor)
[00:31:04] So keep in mind, we're still just on the first inequality

Guest (Customer)
[00:31:09] got it

Eduardo d (Tutor)
[00:31:50] Now, notice, the -3.5 and -3 divides the number line to three intevals:

Guest (Customer)
[00:32:21] got it

Eduardo d (Tutor)
[00:33:03] So, we need to test each of these intervals, and see which gives < 0
[00:35:04] Still with me?

Guest (Customer)
[00:35:07] yes
[00:35:10] working on it right now

Eduardo d (Tutor)
[00:35:24] All right

Guest (Customer)
[00:36:30] all of them gave me less than 0

Eduardo d (Tutor)
[00:36:56] Can you check on them again?

Guest (Customer)
[00:37:19] I plugged in -4, -3.2, and -2

Eduardo d (Tutor)
[00:37:55] Ok, and what did you get for each?

Guest (Customer)
[00:38:09] oh my bad
[00:38:13] the first one is positive :p
[00:38:27] and so is the seocnd one
[00:39:16] ah one second
[00:40:08] how did you get positive for (- infinity, -3.5)?
[00:40:37] i'm confusing myself, sorry,. i plugged it into a+3 instead

Eduardo d (Tutor)
[00:41:09] Ok, make sure to plug it to the whole 2a+7/2(a+3)

Guest (Customer)
[00:42:24] ok
[00:42:25] got it

Eduardo d (Tutor)
[00:43:08] So for the 1st inequality, the solution is (-3.5, -3)

Guest (Customer)
[00:43:19] ok

Eduardo d (Tutor)
[00:43:47] Now, if you do the similar steps to the other inequality, you'll actually end up with (-3, -2.5)

Guest (Customer)
[00:44:02] okay, let me write it down, one sec

Eduardo d (Tutor)
[00:44:25] All right
[00:46:01] I'm working on this...

Guest (Customer)
[00:46:07] okay

Eduardo d (Tutor)
[00:46:28] Sorry, please ignore the auto-status

Guest (Customer)
[00:46:38] oh haha

Eduardo d (Tutor)
[00:46:54] So are you done taking notes?

Guest (Customer)
[00:47:03] almost done
[00:47:30] so
[00:47:38] for values smallter than -3, it's negative

Eduardo d (Tutor)
[00:48:18] Are you on the 2nd inequality?

Guest (Customer)
[00:48:24] yes
[00:52:29] o_o

Cameron S (Tutor)
[00:52:39] Hi there.

Guest (Customer)
[00:52:41] hello

Cameron S (Tutor)
[00:52:50] Looks like you've been transfered?

Guest (Customer)
[00:52:51] can you see anything on the board?
[00:52:53] yes

Cameron S (Tutor)
[00:53:09] Yes, it was flickering, so I didn't know if there was a technical problem

Guest (Customer)
[00:53:20] okay
[00:53:25] it's a really long problem

Cameron S (Tutor)
[00:53:31] I see.
[00:53:38] Do you just want me to check it over?

Guest (Customer)
[00:53:40] do you want me to explain to you where i got everything?

Cameron S (Tutor)
[00:54:01] Looks like you took the derivative first.

Guest (Customer)
[00:54:15] yes, set it to 0 and got

Cameron S (Tutor)
[00:54:27] OK

Guest (Customer)
[00:54:45] since i'm looking for sine, i restricted the function to its range
[00:55:13] from -1 to 1, but since we're solving a for the function where it has no critical numbers

Cameron S (Tutor)
[00:55:34] OK

Guest (Customer)
[00:55:36] i have to set the to less than - 1 and greater than 1
[00:56:04] i solved for a and got -5/2>a and -7/2>a

Cameron S (Tutor)
[00:56:05] Yes, since that equality cannot hold for the answer.

Guest (Customer)
[00:56:13] yes
[00:56:24] but my teacher insisted that we must do a signs table..

Cameron S (Tutor)
[00:56:34] OK

Guest (Customer)
[00:57:08] i solved the 2 inequalities and did the signs tables for both of them
[00:57:28] and for the inequality that's less than -1, I got the range (- 3.5,-3)
[00:57:42] and for the one greater than 1, I got (-3,-2.5)

Cameron S (Tutor)
[00:58:02] OK. Do you have the answer too, or do you want me to check your work?

Guest (Customer)
[00:58:18] well, he said the answer is (- 3.5, -2.5)
[00:58:37] does that mean when you combine the 2 tables, you get a restricted range of (-3,-2.5)?
[00:58:49] i'll draw the second table just to make sure

Cameron S (Tutor)
[00:58:56] That' makes sense because your two answers are OR'd together.
[00:59:28] ...if 'a' satisfies either inequality, it's good.

Guest (Customer)
[00:59:37] okay
[00:59:46] i just hope it makes sense

Cameron S (Tutor)
[00:59:54] One question though...
[01:00:03] You've got open intervals...
[01:00:10] so the answer wouldn't include 3.

Guest (Customer)
[01:00:15] hmm..

Cameron S (Tutor)
[01:00:19] Sorry...
[01:00:36] -3

Guest (Customer)
[01:00:38] so does the range of sine include -1 and 1?
[01:01:16] if it does, wouldn't it make sense to make the inequalities less than or equual to and greater than or equal to?

Cameron S (Tutor)
[01:01:21] Yes, it does, so it makes sense that you used open intervals.

Guest (Customer)
[01:01:53] hmm..

Cameron S (Tutor)
[01:02:00] We want a to make it outside the range of [-1 to 1 inclusive]

Guest (Customer)
[01:02:18] that?

Cameron S (Tutor)
[01:03:32] Not sure what you mean by "that?"

Guest (Customer)
[01:04:01] ifequal to were included in there, it'd make sense, and we wouldn't exclude -3

Cameron S (Tutor)
[01:04:23] I see what you're saying, yes.

Guest (Customer)
[01:04:25] but since the range includes both -1,1, it wouldn't make sense to do so

Cameron S (Tutor)
[01:04:32] Right.
[01:04:48] Can I double check some of your work.
[01:04:49] ?

Guest (Customer)
[01:05:05] do you want me to write everything out?
[01:05:12] or are you working on it?

Cameron S (Tutor)
[01:05:16] No, I think I can do it as is.

Guest (Customer)
[01:05:19] okay

Cameron S (Tutor)
[01:05:23] cool

Guest (Customer)
[01:05:24] go ahead
[01:06:48] wait
[01:07:27] ah nvm. I thought if we were to find the critical points, it'd make sense to have the derivative = 0 as well as where it is undefined

Cameron S (Tutor)
[01:07:59] Yes, we have to think about undefined too.

Guest (Customer)
[01:08:04] so
[01:08:06] for both inequalities

Cameron S (Tutor)
[01:08:11] But...I got different intervals than you.

Guest (Customer)
[01:08:13] if you plug in 3, you'd get an undefined answer
[01:08:18] which makes -3 included in the range
[01:08:22] -3*

Cameron S (Tutor)
[01:08:48] -3 makes it undefined, yes.
[01:08:54] So, that's included.

Guest (Customer)
[01:08:57] so wouldn't it be included then?

Cameron S (Tutor)
[01:09:04] Yes, it would!

Guest (Customer)
[01:09:11] so then for the first one it'd be (-3.5,-3]
[01:09:16] second one as well

Cameron S (Tutor)
[01:09:29] Actually, I got something different from your interval.
[01:09:34] Let me see what you have on the board.

Guest (Customer)
[01:09:37] ok

Cameron S (Tutor)
[01:11:32] Hmm...OK
[01:11:45] I don't see anything wrong with your method.

Guest (Customer)
[01:12:00] so using the undefined to include -3 would be okay?

Cameron S (Tutor)
[01:12:05] Yes

Guest (Customer)
[01:12:08] okay

Cameron S (Tutor)
[01:12:12] Nice work.

Guest (Customer)
[01:12:17] thanks for your help
[01:12:20]

Cameron S (Tutor)
[01:12:28]
[01:12:31] No problem.