MedVision ad

Question help in regards to integration inequality (1 Viewer)

bennyngo12345678

New Member
Joined
Jan 17, 2015
Messages
4
Gender
Male
HSC
2015
help101.JPG

This question has really stumped me in terms of where to even begin. I would really appreciate help for this question.

thank you!!!
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
how did you know where to begin from?
You could also do this the other way round, by using differentiation instead of integration.

i.e. show that , and then show that .

To show the first one, you define the LHS as f(x), show that f(0) = 0, and then show that f′(x) > 0 for all positive x. This means the LHS is 0 at x = 0, and is increasing thereafter, so the LHS is always strictly greater than 0 when x > 0, as required. Then do a similar thing with the second inequality.

So you'll end up having to show the same inequalities as BombMacBomb, but they won't seem like you had to know where to start, they'll appear out of your differentiation.
 

BombMacBomb

New Member
Joined
Apr 9, 2013
Messages
14
Location
Upstairs
Gender
Male
HSC
2010
Uni Grad
2013
how did you know where to begin from?
Well I just took the inequality and worked backwards, differentiating each chunk before splitting the inequalities and proving each side seperately.
For most inequality questions, I usually take the answer and work backwards first. Works brilliantly most of the time.
 

psyc1011

#truth
Joined
Aug 14, 2014
Messages
174
Gender
Undisclosed
HSC
2013
how did you know where to begin from?
His working out is essentially read backwards. Read from bottom to top, that is how he found the starting point. This technique is taught in discrete maths at UNSW (first year course for compsci/SFE/math programs).

He begins from the answer and works backwards to where a fact is met. Then he reverses his working out and it should look ingenious because the reader will ponder on how he thought to start from that inequality and manipulate such that he obtains the result
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top