Calculus Inequality - 2009 JRAHS TRIAL (1 Viewer)

[asianese]

4c
i) If , where c>0, k>0, k=/=1, show y has a single stationary value between x=0 and x=c, and show that this stationary value is a maximum if 0<k<1 and a minimum if k>1. (3)

ii) Hence show that if a>0, b>0, a=/=b, then if 0<k<1 and if k>1 (4)

I've done i)
for stationary points, ends up

For k<1, k-1<0, so cc down, rel max. For k>1, k-1>0, so cc up, rel min.

ii)

At
If maximum, then
If minimum,...

Don't know where to go

Any help is greatly appreciated.

 

[funnytomato]

just for the 1st part of ii)

compare the given inequality with the one you obtained in part i):
y= [ x^k-(c-x)^k ] /2 <= (c/2)^k ,
we'll need to make the following substitutions :
let c=a+b
then we can let x=a, so c-x=a+b-a=b
and we get a strict inequality since:
a=/=b, so x=/= c/2
but x=c/2 is the only point where MAX/MIN (hence equality) is achieved

and the 2nd part of ii) is quite similar

 

[asianese]

I'm guessing you're latexing atm

OMG THAT'S COOL! llol thanks so much

and yeah I thought of doing that but I just wasn't brave enough. thanks!!

 

[funnytomato]

I'm guessing you're latexing atm

sorry, my bad
they're written in white , you gotta highlight them to read

I didn't wanna spoil it, so before you read it, these are some hints:
basically you need to compare the inequaity you can obtain from part i)
and the one to prove in part ii)
Then you'll need to make some substitutions

it's not as difficult as you may think it is

 

[funnytomato]

I'm guessing you're latexing atm

OMG THAT'S COOL! llol thanks so much

and yeah I thought of doing that but I just wasn't brave enough. thanks!!

and you can check that the restrictions are satisfied , when you substitue