Tutor.com The Derivative Session

Jan. 21, 2014

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Session Transcript - Math - Calculus, 1/5/2014 3:42PM - Tutor.comSession Date: 1/5/2014 3:42PM
Length: 42.4 minute(s)
Subject: Math - Calculus


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

Casey (Customer)
[00:00:00] The first derivative of the function f is given by f'(x)=(cos^2x)/x - (1/5) . How many critical values does f have on the open interval (0,10)? [ File > http://lhh.tutor.com/SharedSessionFiles/a98bcbe6-5d1d-4bb7-9083-b1e2b8604ab7_image_(16).jpg ]

Andrew S (Tutor)
[00:00:16] Hi :)

Casey (Customer)
[00:00:37] hi, i'm stuck on these two questions!
[00:00:57] i'm not sure if i'm on the right track for either :/

Andrew S (Tutor)
[00:01:28] So, for the first one, did you find what f(x) is?

Casey (Customer)
[00:01:56] f(x) is sin^2x-(1/5x)+c i think

Andrew S (Tutor)
[00:02:42] Well, so first of all, that isn't the right anti-derivative, and second, you don't need to find what f(x) is

Casey (Customer)
[00:03:03] oh.. okay.. what do i need to do?

Andrew S (Tutor)
[00:03:11] Remember that f(x) will have a critical value whenever f'(x) is 0
[00:03:24] so what you need to do is set f'(x)=0 and solve
[00:03:36] remember that trig values repeat themselves
[00:03:46] Does that make sense?

Casey (Customer)
[00:03:50] yes

Andrew S (Tutor)
[00:04:28] Why don't we start on a less crowded page

Casey (Customer)
[00:04:34] ok

Andrew S (Tutor)
[00:06:08] Hmm, so we don't really need to solve it, as in find the exact answers, we just need to find out how many their are
[00:06:12] *there

Casey (Customer)
[00:06:34] how do i find that?
[00:07:14] oh would it be three bc it would be 1/x set equal to zero and then cos^2 would get 2 answers?

Andrew S (Tutor)
[00:07:38] Well, a really easy way, if you have a graphing calculator, is to just graph it and count the number of times it crosses the x-axis between 0 and 10

Casey (Customer)
[00:08:35] this question is in the non calculator section.. but can i write as an explanation that critical values of f are found when f'(x) is set equal to zero ?

Andrew S (Tutor)
[00:09:09] Oh, well you are right the answer is 3 :)

Casey (Customer)
[00:09:19] ok!

Andrew S (Tutor)
[00:09:24] And that is a good explanation
[00:10:14] So does that all make sense for number 7? The answer should be three, you understand everything about how we go it?

Casey (Customer)
[00:10:51] yes thank you! do i need to upload the picture again so you can see #8 or is it still here somewhere?

Andrew S (Tutor)
[00:11:00] It's still there
[00:11:14] if you go back to the first page. Look in the lower right corner
[00:11:21] you can see the pages we have

Casey (Customer)
[00:11:28] oh! thanks

Andrew S (Tutor)
[00:12:13] So for this next one ... we want to know the line tangent to the graph when f'(x)=1

Casey (Customer)
[00:13:02] so would i have to find the derivative first? to find the slope of the tangent line?

Andrew S (Tutor)
[00:13:29] Well, remember this
[00:13:45] f'(x) is the slope of the tangent line
[00:13:57] so if f'(x)=1
[00:14:03] then we know that the slope is 1
[00:14:09] does that make sense so far?

Casey (Customer)
[00:14:12] yep!

Andrew S (Tutor)
[00:14:20] So the answer is not A
[00:15:00] Now, if we wanted to, we could probably do what you are trying to do, by solving this for x, ect
[00:15:13] however, I think there might be an easier way

Casey (Customer)
[00:15:12] ok

Andrew S (Tutor)
[00:15:37] The line tangent to the graph at that point has to touch the graph at that point
[00:15:50] So when y=1
[00:16:12] the x values have to agree as well
[00:16:37] So what we can do is solve B, C, D, and E for x, when y=1
[00:16:41] and plug that in f(x)
[00:16:48] and see if we get 1
[00:16:51] does that make sense?

Casey (Customer)
[00:18:01] um..so when y=1, x has to = 1 too?

Andrew S (Tutor)
[00:18:45] No, but I let's see if my explanation makes sense with a picture
[00:19:32] I'm not entirely what the graph of f(x) looks like, but something like that
[00:19:40] so at y=1 we get a tangent line
[00:20:07] Something like that

Casey (Customer)
[00:20:17] ok...

Andrew S (Tutor)
[00:20:30] These have to agree at this point
[00:20:47] both of f(x) and the tangent line have to agree at that point

Casey (Customer)
[00:20:55] ok

Andrew S (Tutor)
[00:21:03] We know the y value of that point, it is 1
[00:21:35] So, we are given four possible linear equation
[00:21:40] *equations
[00:21:52] It is really really simple to solve these linear equations for x
[00:22:00] so what I'm suggesting we do is solve them for x
[00:22:17] Oh! you can't you a calculator ...

Casey (Customer)
[00:22:24] yep :(

Andrew S (Tutor)
[00:22:41] I'm sorry, with a calc that would have saved us a lot of time

Casey (Customer)
[00:23:45] it's ok! so if i was to do this by hand, should i find the derivative of f(x)=x^4+2x^2 , then set it = to 1 and solve for x?

Andrew S (Tutor)
[00:24:23] Let's try that
[00:26:01] Wait, hold on

Casey (Customer)
[00:26:03] what do i do now?

Andrew S (Tutor)
[00:26:15] if you can't use a calculator, how are you suppose to get that whole decimal thing
[00:26:27] Those are some long decimals, doing that by hand would be hard

Casey (Customer)
[00:27:43] i'm not sure.... my teacher said this part is non calc and gave us the answers, but we just have to show all the work.. so i know the answer is D but idk how to get it without a calculator...
[00:27:58] maybe like you said before to test all the answers?

Andrew S (Tutor)
[00:28:30] Well, that would be really hard because you would have to raise all of those decimals to some power

Casey (Customer)
[00:29:00] oh...

Andrew S (Tutor)
[00:29:42] I'm sorry, I'm not sure if it is really doable without a calculator ...
[00:29:50] Because of those decimals

Casey (Customer)
[00:30:18] okay.. i guess i'll just use a calculator and tell her i couldn't do it without one

Andrew S (Tutor)
[00:31:10] So if we solve for x :) with a calculator, I get x=0.237

Casey (Customer)
[00:32:03] did you do f'(x)=4x^3 + 4x =1 to get that?

Andrew S (Tutor)
[00:32:20] Right,
[00:32:32] That is the equation I solved for x

Casey (Customer)
[00:32:57] ok, i got .237 too!

Andrew S (Tutor)
[00:33:10] Alright :)
[00:34:16] So we use that to find the intercept
[00:34:18] like I did
[00:34:23] and that is what we get
[00:34:26] does that all make sense?

Casey (Customer)
[00:34:30] yes

Andrew S (Tutor)
[00:34:38] Alright :)
[00:34:41] questions on it?

Casey (Customer)
[00:35:27] i understand how you did this... but the answer is supposed to be D...according to my teacher's answer key :(
[00:35:32] maybe she's wrong?

Andrew S (Tutor)
[00:35:58] Hmmm, let me check some things and see ...

Casey (Customer)
[00:36:31] ok

Andrew S (Tutor)
[00:38:05] I'm getting a little confused now ...

Casey (Customer)
[00:38:18] oh no...
[00:38:34] idk she could be wrong..!

Andrew S (Tutor)
[00:38:43] No, she isn't

Casey (Customer)
[00:38:53] oh..

Andrew S (Tutor)
[00:38:56] I graphed f(x)
[00:39:00] and each of the lines
[00:39:03] on my calc
[00:39:07] and the right one is D

Casey (Customer)
[00:39:10] hm...
[00:39:25] okay....
[00:40:02] for my work should i write down that i graphed everything and draw the graph i see on my calc?

Andrew S (Tutor)
[00:40:17] That seems like a good diea
[00:40:20] idea
[00:40:50] I'm sorry, I'm not sure why my last method didn't work

Casey (Customer)
[00:41:08] it's ok!

Andrew S (Tutor)
[00:41:43] Well there you go :) you may want to talk to your teacher on how you should do it without a calc since I'm not sure how you would
[00:41:56] Especially with those decimals :(

Casey (Customer)
[00:42:08] okay sounds good!

Andrew S (Tutor)
[00:42:16] Have a good day :)

Casey (Customer)
[00:42:22] thanks you too! :)