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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) = (aÂ² + 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) aÂ²+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.