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System Message

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

Jamal (Customer)

[00:00:00] I need help solving b, c, and d. [ File > http://lhh.tutor.com/SharedSessionFiles/4eab3219-c696-495a-afa9-8793f671b911_2012-12-11_164223.pdf ]

Marcelo G (Tutor)

[00:00:04] Hello!

[00:00:08] Welcome to tutor.com ! :)

[00:00:12] How are you today?

Jamal (Customer)

[00:00:20] hello, i am good :)

Marcelo G (Tutor)

[00:00:32] ThatÂ´s nice to read :)

[00:00:35] How can I help you?

Jamal (Customer)

[00:00:55] I need help going through problems b c and d

Marcelo G (Tutor)

[00:01:04] ok

[00:01:12] thus, youÂ´ve already solved a, right?

Jamal (Customer)

[00:01:28] yes, that is correct

Marcelo G (Tutor)

[00:01:33] ok

[00:01:41] we have the derivative of f given by

[00:02:13] and the problem asks for the x coordinate of the critical points, right?

Jamal (Customer)

[00:02:25] yes

Marcelo G (Tutor)

[00:02:43] there is a problem though

Jamal (Customer)

[00:03:10] what would that be?

Marcelo G (Tutor)

[00:03:15] because the function, nor the derivative are defined at x=0 right?

[00:03:37] in both cases we have a zero denominator at x=0

Jamal (Customer)

[00:03:41] oh yes, so it wouldnt be
0 then

Marcelo G (Tutor)

[00:03:56] right :)

[00:04:19] moreover

[00:04:33] the problem says the function as for itÂ´s domain all the real positive numbers, right?

[00:04:40] has*

Jamal (Customer)

[00:04:58] oh yes

Marcelo G (Tutor)

[00:05:15] thus, we shouldnÂ´t consider negative values of x either

[00:05:17] does it make sense?

[00:05:43] great :)

Jamal (Customer)

[00:05:41] yes, so would a critcal number be sqrt2>?

Marcelo G (Tutor)

[00:05:55] right :)

[00:06:23] and we need to find out if it is a local minimum, maximum or neither...

Jamal (Customer)

[00:06:23] ok, and how can we test if it is a relative minimum max or neither

Marcelo G (Tutor)

[00:06:28] ok

[00:06:47] one way is to check the signs of the derivative to the left and to the right of the critical number

[00:07:23] in order to do that we can choose two simple values to the left and right of the critical value

[00:07:29] for example we can choose x=1 and x=2

Jamal (Customer)

[00:07:36] ok, so 1 and 3

[00:07:39] or 2\

[00:08:08] and do we plug in those numbers back into the first equation or into the derivative?

Marcelo G (Tutor)

[00:08:23] we need to plug them into the derivative

[00:08:43] because the sign of the derivative there will tell us if the function is increasing or decreasing at each side

[00:09:02] for example

Jamal (Customer)

[00:08:58] ok

Marcelo G (Tutor)

[00:09:07] what do we get for the derivative at x=1?

[00:09:24] good :)

[00:09:31] and at x=2?

Jamal (Customer)

[00:09:55] negative

Marcelo G (Tutor)

[00:10:00] great :)

[00:10:16] that means the original function goes from increasing to decreasing at the critical point, right?

Jamal (Customer)

[00:10:34] yes

[00:10:41] so it would be a max

Marcelo G (Tutor)

[00:10:55] Excellent :)

[00:11:12] do you have any questions up to now?

Jamal (Customer)

[00:11:29] no, i understand it so far :)

Marcelo G (Tutor)

[00:11:40] ok

[00:11:50] next we need to find the intervals where the function is concave up, right?

Jamal (Customer)

[00:12:14] yes

Marcelo G (Tutor)

[00:12:29] any ideas about how can we find that?

Jamal (Customer)

[00:12:45] find the second derivative

Marcelo G (Tutor)

[00:13:00] right :)

[00:13:20] how can we find the second derivative?

Jamal (Customer)

[00:13:33] quotient rule

Marcelo G (Tutor)

[00:13:40] right :)

[00:14:13] good :)

[00:14:24] can we simplify that expression a bit?

Jamal (Customer)

[00:14:40] yes

[00:14:53] is that right?

Marcelo G (Tutor)

[00:15:05] what is the power in the denominator?

Jamal (Customer)

[00:15:20] oh ok it would be 8

Marcelo G (Tutor)

[00:15:26] right :)

Jamal (Customer)

[00:15:39] and the top would be to the 5th, right?

Marcelo G (Tutor)

[00:15:44] because we have (x^a)^b = x^(aÂ·b)

[00:16:03] the top is ok :)

Jamal (Customer)

[00:16:09] ok

Marcelo G (Tutor)

[00:16:16] what can we do next?

[00:16:40] good :)

[00:17:04] is there anything we can factor out in the numerator?

Jamal (Customer)

[00:17:01] one moment please

Marcelo G (Tutor)

[00:17:13] sure :)

Jamal (Customer)

[00:18:32] ok sorry about that

Marcelo G (Tutor)

[00:18:40] no problem :)

Jamal (Customer)

[00:19:09] there we go

Marcelo G (Tutor)

[00:19:17] good :)

[00:19:21] anything we can simplify?

Jamal (Customer)

[00:19:44] can we cancel out the powers?

Marcelo G (Tutor)

[00:19:55] which ones?

Jamal (Customer)

[00:20:06] the 2x^3 and x ^8

Marcelo G (Tutor)

[00:20:18] what would it be?

[00:20:50] great :)

[00:21:01] there we have the second derivative of function f(x)

[00:21:33] now, we want to find the intervals in which the function f(x) is concave up, right?

Jamal (Customer)

[00:21:48] correct

Marcelo G (Tutor)

[00:22:02] how does that relate to the second derivative?

Jamal (Customer)

[00:22:36] if the critcal number is greater than 0, it is concave up i believe. or something like that

Marcelo G (Tutor)

[00:22:57] well, in this case we need to check for the values of the second derivative

[00:23:07] second derivative greater than zero means concave up

[00:23:20] second derivative less than zero means concave down

Jamal (Customer)

[00:23:35] i see

Marcelo G (Tutor)

[00:23:38] as the second derivative in this case is continuous in the domain x>0

[00:23:46] it only can change itÂ´s sign where it is zero

[00:24:06] thus I would first try to find the point or points at which the second derivative is zero

Jamal (Customer)

[00:24:13] ok

Marcelo G (Tutor)

[00:24:55] remember the domain of our original function is values of x greater than zero

[00:25:13] thus we donÂ´t consider negative values of x in this case, right?

Jamal (Customer)

[00:25:12] ok so just positive 2

Marcelo G (Tutor)

[00:25:23] right :)

Jamal (Customer)

[00:25:19] yes

Marcelo G (Tutor)

[00:25:36] at x=2 the second derivative is zero so it can change itÂ´s sign there

[00:25:56] again we need to check for the signs of the second derivative in this case to the left and to the right of x=2

[00:26:12] good :)

[00:26:31] what do we get for the second derivative at x=1?

[00:26:42] right :)

[00:26:54] that means to the left of x=2 the graph is concave down, right?

Jamal (Customer)

[00:27:13] yes

Marcelo G (Tutor)

[00:27:33] what do we get for the second derivative at x=3?

[00:27:42] great :)

Jamal (Customer)

[00:27:42] positive

Marcelo G (Tutor)

[00:27:50] thus to the right of x=2 the graph is concave up

[00:28:00] how can we write the interval in which it is concave up?

Jamal (Customer)

[00:28:12] so the interval would be (2, inf)

Marcelo G (Tutor)

[00:28:19] perfect! :)

[00:28:31] Do you have any questions up to now?

Jamal (Customer)

[00:28:59] no i understand it all so far

Marcelo G (Tutor)

[00:29:08] ok

[00:29:33] next the problem asks for the value of x where the line tangent to the graph is parallel to the line y=x right?

Jamal (Customer)

[00:29:52] correct

Marcelo G (Tutor)

[00:30:11] what is the condition that must be satisfied for two lines to be parallel?

Jamal (Customer)

[00:30:27] they must have the same slope, or derivative.

Marcelo G (Tutor)

[00:30:41] same slope right :)

[00:30:46] what is the slope of the line y=x?

Jamal (Customer)

[00:30:52] 1

Marcelo G (Tutor)

[00:31:07] good :)

[00:31:28] the slope of the line tangent to the graph at a given values of x is given by the value of the derivative, right?

Jamal (Customer)

[00:31:45] yes

Marcelo G (Tutor)

[00:32:13] thus, this amounts to find the value or values of x such that the derivative is 1, does that make sense?

Jamal (Customer)

[00:32:31] oh i get it :)

Marcelo G (Tutor)

[00:32:49] how can we set it up?

Jamal (Customer)

[00:33:04] like that?

Marcelo G (Tutor)

[00:33:10] great :)

[00:33:14] can we solve that for x?

[00:33:55] well, that last step is not correct

Jamal (Customer)

[00:33:59] oh

Marcelo G (Tutor)

[00:34:02] because we have an addition there, right?

Jamal (Customer)

[00:34:11] yes

Marcelo G (Tutor)

[00:34:22] we have

[00:34:37] right?

Jamal (Customer)

[00:34:38] oh i see

Marcelo G (Tutor)

[00:34:52] this is not a simple equation to solve

Jamal (Customer)

[00:35:14] is thar what we doe?

Marcelo G (Tutor)

[00:35:30] yes, in this case we are trying to find one solution

[00:35:41] let me rewrite it like this

[00:35:59] is that ok?

Jamal (Customer)

[00:36:05] yes

Marcelo G (Tutor)

[00:36:27] as a quartic equation as the one we have is not an easy task to solve

[00:36:39] what we can do is to try with simple integers to see if they work

[00:36:51] Any ideas?

Jamal (Customer)

[00:36:47] ok

[00:37:02] 1

Marcelo G (Tutor)

[00:37:07] right :)

[00:37:11] x=1 works :)

[00:37:31] let me check if there are others

Jamal (Customer)

[00:37:38] ok

Marcelo G (Tutor)

[00:37:57] no, there arenÂ´t :)

[00:38:03] at least in the domain of our function

[00:38:08] thus our answer is x=1

[00:38:16] does it make sense?

Jamal (Customer)

[00:38:35] yes :)

Marcelo G (Tutor)

[00:38:46] Do you have any questions about it?

Jamal (Customer)

[00:39:07] thanks so much for all the help, i appreciate it :) no i dont have any questions

Marcelo G (Tutor)

[00:39:28] you did a great job, keep on that track!! :)

Jamal (Customer)

[00:39:46] thanks, have a good evening, bye!

Marcelo G (Tutor)

[00:39:57] Thank you, you too have a good evening :)

[00:39:59] Good bye ;)