Proof Help Needed

[h2p]

x = 3p^2 − 2p^3

Show that x < p iff p < 1/2

note: p^2 = p'squared and p^3 = p'qubed

Thanks!

 

[Unco]

That is not true. Try plugging in p=0 or any negative number to show x>p when p < 0.

Edit: corrected x>p.

 

[h2p]

I should have mentioned this is from a probability example so the value that p can take on is between 0 and 1. So basically between 0 as a lower limit and less than 0.5 for the upper limit is should hold true. Unfortunatly plugging in number doesnt act as a proof

 

[Denis]

Not sure what you're trying to accomplish, h2p (should that be h2o?!).

If x = 0, then p = 0 or p = 3/2 ....

 

[h2p]

I need to prove (without plugging numbers in) that the equation holds true for the range of p specified.

 

[Unco]

x = 3p² − 2p³

If x > p,

3p² - 2p³ < p

2p³ - 3p² + p > 0

p(2p - 1)(p - 1) > 0

Solutions if it was an equality: p=0, p=1/2, p=1.

We know that cubic is of type up-down-up, so it is above the x-axis in
0 < p < 1/2.