Insight 2008 MM Exam1 Trial Error

[pHysiX]

Question 1 b):

the answer says log_e|x-3|+4

however, if we look at the derivative function's domain, we see that it's (-infinity, 3). Therefore, the answer should be the negative reflection of the log, i.e.

log_e(3-x)+4.

Please tell me if you agree with me because this type of question appeared in 2006 Sample Exams and they clearly solved it like that. =]

[ReVeL]

Wouldn't happen to have a version you could upload?

I've only got insight 2008 exam 2...

[danieltennis]

Godspeed.

[danieltennis]

Solutions

[shinny]

I don't think its necessarily an error, nor would you lose a mark for doing writing what they did, nor what you did; it's purely just a technicality. The domain restriction acts to give the answer you gave, but keeping the modulus has no difference in terms of the graph that is produced if you get what I mean, since the domain restriction still stays and produces what seems like .

[pHysiX]

mmm i agree with you there. but wen u look @ 2006 Sample Exam, I believe it's better to consider the other answer because it demonstrates that you understand functions etc. because really, the exams are about demonstrating knowledge.

[riadnicolas]

graphing the function might help you understand, compare gradient of modulus function and the derivative

[shinny]

Its the equivalent of cases where you have say, . The modulus is UNNECESSARY because the inside will always be greater than 0, but that doesn't mean it's wrong. Same with the case above; although the domain restriction implies that the inside of the modulus will always be negative, it doesn't mean its wrong to keep it there. But as you said, I would remove the modulus too and put the 'more correct' version just as insurance.

[Mao]

Question 1 b):

the answer says log_e|x-3|+4

however, if we look at the derivative function's domain, we see that it's (-infinity, 3). Therefore, the answer should be the negative reflection of the log, i.e.

log_e(3-x)+4.

Please tell me if you agree with me because this type of question appeared in 2006 Sample Exams and they clearly solved it like that. =]

I gave the answer like you did, and you are correct.  However as other people have said, there is also a domain restriction.  This makes the modulus log function also correct.  On a VCAA exam, I'd write it how you did OR use the modulus log, however re-write the domain of the function, next to your answer, so they know you've limited it [as required in the question].

P.S. This isn't Mao.

[Glockmeister]

P.S. This isn't Mao.

Huh? Has he finally manage to divide by zero?

[amycourty]

would anyone happen to have the insight 2008 exam 2 and be willing to share with me?  :angel:

[Chocoholic]

jess posted it in this thread
http://vcenotes.com/forum/index.php/topic,6375.15.html

[amycourty]

thanks muchly