Tutor.com Functions Session

Oct. 11, 2013

Session Transcript - Math - Calculus, 9/25/2013 10:16PM - Tutor.comSession Date: 9/25/2013 10:16PM
Length: 42 minute(s)
Subject: Math - Calculus

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

Keven (Customer)
[00:00:00] finding domain

Sarvnaz F (Tutor)
[00:00:08] Hello

Keven (Customer)
[00:00:10] Hi!

Sarvnaz F (Tutor)
[00:00:12] Welcome to tutor.com

Keven (Customer)
[00:00:40] I need help finding the domain of a function. Would you like me to draw it up there?

Sarvnaz F (Tutor)
[00:02:17] Ok, so do you have any idea where to begin here?

Keven (Customer)
[00:02:43] finding restrictions?

Sarvnaz F (Tutor)
[00:02:48] yes
[00:02:55] and how would you find that for this function?

Keven (Customer)
[00:03:20] set them equal to zero?

Sarvnaz F (Tutor)
[00:03:23] Do you know what is the restriction for square root?

Keven (Customer)
[00:03:32] cant be negative

Sarvnaz F (Tutor)
[00:03:35] true
[00:03:55] So, another way to represent that is it has to be greater than 0
[00:03:58] Are you with me?

Keven (Customer)
[00:04:04] Yes

Sarvnaz F (Tutor)
[00:04:34] good. So, the first thing we need to do for the numerator is to set the expression under the square root >-0
[00:04:43] I meant >=0
[00:04:54] because in the numerator, we can have 0
[00:04:57] Can you do that?

Keven (Customer)
[00:05:22] Not really, I dont understand that well, can you show me?

Sarvnaz F (Tutor)
[00:05:28] yes
[00:05:59] x^2+3 >=0
[00:06:07] This will give us an equation to solve for
[00:06:12] Any idea how to solve this for x?

Keven (Customer)
[00:06:52] I think thats wrong

Sarvnaz F (Tutor)
[00:06:59] no it is right
[00:07:37] But if x^2=-3, that means x=squrt(-3) which is not a real number

Keven (Customer)
[00:07:43] imaginary

Sarvnaz F (Tutor)
[00:07:57] Right, Hence, this means x^2 can never =-3
[00:09:12] and x^2 is always >-3 for all real numbers
[00:09:18] So the answer here is all real numbers
[00:09:22] Are you following me?

Keven (Customer)
[00:09:27] yes

Sarvnaz F (Tutor)
[00:09:36] ok. So this is for the numerator
[00:09:45] we have to do the same for the denominator
[00:10:02] Are you with me?

Keven (Customer)
[00:10:05] yes

Sarvnaz F (Tutor)
[00:10:16] What are the restrictions for the denominator?

Keven (Customer)
[00:10:29] cant equal 0

Sarvnaz F (Tutor)
[00:10:33] true
[00:10:51] and in this case, since the denominator is also under the radical, it has to also be >0
[00:11:01] So we have to set x^2>0
[00:11:29] So, for x^2>0
[00:11:51] this expression will always be greater than zero
[00:11:55] Do you understand why?

Keven (Customer)
[00:12:16] yes because there cant be a negative
[00:12:26] and cant equal to zero

Sarvnaz F (Tutor)
[00:12:52] yes, when you square any number, negative or positive, the result is always a positive number
[00:13:04] So x^2 is always >0

Keven (Customer)
[00:13:04] Yes

Sarvnaz F (Tutor)
[00:13:35] So, the only restriction for domain here is x can't equal 0

Keven (Customer)
[00:13:47] okay

Sarvnaz F (Tutor)
[00:14:09] So, any idea how to write the domain now?
[00:15:00] that is half of the answer
[00:15:09] we can have negative numbers here too
[00:16:33] So, that is the domain
[00:16:43] Are you following how I got that?

Keven (Customer)
[00:16:49] my teacher said the answer was x cannot equal (+ or -) 5(sq root of 2)/2

Sarvnaz F (Tutor)
[00:17:35] Oh, I see
[00:18:03] I had a typo when I transfered the solution over

Keven (Customer)
[00:18:15] ah okay

Sarvnaz F (Tutor)
[00:18:27] The denominator can't =0, we have to set the whole denominator =0

Keven (Customer)
[00:19:05] okay

Sarvnaz F (Tutor)
[00:19:41] When I read the problem, I thought it was 5*sqrt(2x^2)

Keven (Customer)

Sarvnaz F (Tutor)
[00:20:13] So, now do you know how to solve this equation?

Keven (Customer)
[00:21:01] no, not since we had to change the equation up because of the misread

Sarvnaz F (Tutor)
[00:21:18] ok, let me get a new board
[00:21:34] Actually, let me see if I can do it here

Keven (Customer)
[00:21:39] okay

Sarvnaz F (Tutor)
[00:21:57] So, like you said earlier, the denominator restriction is that it can't equal =0
[00:22:05] So, the denominator for this equation can't equal =

Keven (Customer)
[00:22:06] right

Sarvnaz F (Tutor)
[00:22:39] So, we need to set the denom not=0, and solve for x
[00:23:08] To solve this, we can subtract 5 from both sides
[00:24:06] Then we square both sides to get rid of square root
[00:24:36] Divide both sides by 2
[00:25:14] You with me so far?

Keven (Customer)
[00:25:17] yex
[00:25:18] yes

Sarvnaz F (Tutor)
[00:25:27] Now, we take square root of both sides
[00:25:42] What is square root of 25?

Keven (Customer)
[00:25:46] 5

Sarvnaz F (Tutor)
[00:26:18] Now, we need to rationalize by multiplying numerator and denominator by square root of 2
[00:27:08] That is how you get the answer

Keven (Customer)
[00:27:11] Ah okay
[00:27:16] I got you! Thank you

Sarvnaz F (Tutor)
[00:27:22] no problem

Keven (Customer)
[00:27:29] Is there someone waiting for you?

Sarvnaz F (Tutor)
[00:27:45] No

Keven (Customer)

Sarvnaz F (Tutor)
[00:27:47] Why?

Keven (Customer)
[00:27:58] may i do another problem with you?

Sarvnaz F (Tutor)
[00:28:05] sure. I can do one more

Keven (Customer)
[00:28:22] okay here comes a hard one
[00:28:29] new board?

Sarvnaz F (Tutor)
[00:29:10] there

Keven (Customer)
[00:29:13] okay lets get started
[00:29:16] thank you

Sarvnaz F (Tutor)
[00:29:21] sure

Keven (Customer)
[00:30:56] There we are

Sarvnaz F (Tutor)
[00:30:59] ok
[00:31:04] Any idea how to do this problem?

Keven (Customer)
[00:31:17] not at all

Sarvnaz F (Tutor)
[00:31:25] Ok, fog
[00:31:30] can be written like this
[00:31:47] Are you with me?

Keven (Customer)
[00:31:53] yes

Sarvnaz F (Tutor)
[00:32:05] now, instead of g(x), we plug the function
[00:33:04] So, now that we did this, in the function of f(x), anywhere we have x, we have to plug (x/5 +4) instead
[00:33:08] does that make sense?

Keven (Customer)
[00:33:11] yes

Sarvnaz F (Tutor)
[00:33:15] can you do that?
[00:33:34] Give it a try on the board, I will walk you through it if you get stuck
[00:34:09] yup
[00:34:12] You got it
[00:34:16] Now we need to simplify this
[00:34:24] any idea how to do that?
[00:35:16] yup

Keven (Customer)

Sarvnaz F (Tutor)
[00:35:45] gof would be the other was
[00:36:16] very close

Keven (Customer)
[00:36:21] that would be the answer

Sarvnaz F (Tutor)
[00:36:24] 4 is not under the square root
[00:36:31] yup
[00:36:36] This will be the answer

Keven (Customer)
[00:37:06] okay
[00:37:16] and one more thing
[00:37:21] if you have time

Sarvnaz F (Tutor)
[00:37:30] sure. just one more
[00:37:37] I have to go after that

Keven (Customer)
[00:37:40] Ok i promise this will be the last one

Sarvnaz F (Tutor)
[00:37:48] ok. here is a new board

Keven (Customer)
[00:38:02] find a linear model that best fit points
[00:39:15] okay

Sarvnaz F (Tutor)
[00:39:29] ok, do you need to use a calculator for this?

Keven (Customer)
[00:39:40] i could if i wanted to
[00:39:45] would it be best?

Sarvnaz F (Tutor)
[00:39:57] What method has your teacher asked you to use for this?

Keven (Customer)
[00:40:14] the table function on our calculator

Sarvnaz F (Tutor)
[00:40:30] yes, then you need to use your calculator

Keven (Customer)
[00:40:40] okay ill be back in 10 sec
[00:41:06] ok
[00:41:09] i got it

Sarvnaz F (Tutor)
[00:41:12] ok. my shift is ending. you might want to log back in and ask another tutor to help you.

Keven (Customer)
[00:41:21] okay

Sarvnaz F (Tutor)

Keven (Customer)
[00:41:28] thank you for everything

Sarvnaz F (Tutor)
[00:41:32] no problem

Keven (Customer)
[00:41:37] and its totally okay you helped me a lot

Sarvnaz F (Tutor)
[00:41:39] thanks for your patience and using tutor.com
[00:41:52] have a great rest of the week

Keven (Customer)
[00:41:57] you too!!

Sarvnaz F (Tutor)
[00:42:01] Bye for no