Tutor.com The Derivative Session

Mar. 15, 2012

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Session Transcript - Math - Calculus, 3/10/2012 8:43PM - Tutor.comSession Date: 3/10/2012 8:43PM
Length: 17.6 minute(s)
Subject: Math - Calculus


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

Guest (Customer)
[00:00:02] verify that the function satisfies the hypotheses of the mean value thrm on the given interval then find all numbers c that satisfy the conclusion of the mean value thrm. f(x)=x^3-3x+2 [-2,2]

Christine M (Tutor)
[00:00:11] Welcome to Tutor.com ! My name is Christine. Let's get started! :)

Guest (Customer)
[00:00:16] hello

Christine M (Tutor)
[00:00:43] Do you have any ideas on how we'd start this problem?

Guest (Customer)
[00:00:54] um finding critcal numbers?
[00:00:56] I am really not sure

Christine M (Tutor)
[00:01:18] Ok, do you remember what the mean value theorem says?

Guest (Customer)
[00:01:45] not really

Christine M (Tutor)
[00:02:55] Ok, I'm paraphrasing, but it says that given a function between two endpoints (the interval you're given), there is at least one point on the function where the tangent of it is parallel to the line that passes through the end points of the arc
[00:03:07] That's pretty wordy, so I'll draw what I mean
[00:03:33] Let's say this blue arc is our function

Guest (Customer)
[00:03:51] ok

Christine M (Tutor)
[00:04:09] And the black line I drew is the secant line, or the straight line between the two ends of the arc
[00:04:43] The mean value theorem says that there's at least one point on this arc where the tangent is parallel to that secant line
[00:04:45] like so...
[00:05:16] In the case I drew here, there were two points that satisfied that
[00:05:24] Does that make a little more sense?

Guest (Customer)
[00:06:33] yes I see it now

Christine M (Tutor)
[00:06:51] Ok, great! :)
[00:07:10] So we want to find those points with this function f(x) = x^3-3x+1
[00:07:12] +2, sorry
[00:07:23] Between [-2,2]
[00:07:38] The first thing we need to do is find where those endpoints are
[00:07:46] evaluating the function at f(-2) and f(2) to get the y-values

Guest (Customer)
[00:08:02] okay let me do that

Christine M (Tutor)
[00:08:09] :)

Guest (Customer)
[00:08:41] f(-2) is 0

Christine M (Tutor)
[00:08:49] right

Guest (Customer)
[00:08:58] and f(2) is 4

Christine M (Tutor)
[00:09:04] Great!
[00:09:13] So the endpoints are (-2,0) and (2,4)
[00:09:28] Do you remember how to find the slope between two points from algebra?

Guest (Customer)
[00:09:50] y-y/x-x

Christine M (Tutor)
[00:09:59] yep! :)

Guest (Customer)
[00:10:26] :)

Christine M (Tutor)
[00:10:45] Once you calculate that, we can move onto the next step. :D

Guest (Customer)
[00:11:00] let me do so
[00:11:24] 1

Christine M (Tutor)
[00:11:33] Right
[00:11:49] Ok, so now we need to take the derivative of the function and set it equal to that slope (1)
[00:11:54] Then solve for x
[00:12:35] Are you comfortable with derivatives? Not sure how far in your course you are

Guest (Customer)
[00:12:51] yes I can do that
[00:13:01] 3x^2-3

Christine M (Tutor)
[00:13:15] Ok, so...
[00:13:42] Solve for x and we'll have the points where the mean value theorem is met! :)

Guest (Customer)
[00:14:29] let me do so!
[00:14:55] sqrt +/- 4/3

Christine M (Tutor)
[00:15:05] Perfect!
[00:16:10] Those are your two values and the answer to the problem. :D

Guest (Customer)
[00:16:29] thats it???

Christine M (Tutor)
[00:16:36] Yep, that's all
[00:16:38] lol

Guest (Customer)
[00:16:37] wow simple
[00:16:39] thank you so much

Christine M (Tutor)
[00:16:46] No problem!

Guest (Customer)
[00:16:46] Idk what I was thinking
[00:16:47] thank you

Christine M (Tutor)
[00:16:55] Eh, the theorems are wordy
[00:17:01] Sometimes it's hard to make sense of what they mean

Guest (Customer)
[00:16:59] they are
[00:17:03] youre a great tutor!

Christine M (Tutor)
[00:17:12] Have a great evening! If you liked the session, please fill out the survey as you leave! :)

Guest (Customer)
[00:17:20] I will youre great
[00:17:26] thanks

Christine M (Tutor)
[00:17:28] Thanks! :)