solving cubic inequality.. pls help!

[lollipop2046]

2p^3 - 3p^2 - 3p - 1 > 0

is there a way to solve this algebraically, just simply without the help of claculator?

 

[stapel]

Since the polynomial has no rational roots, I don't see how you'd be able to factor, which would be the usual way to proceed "algebraically". Numerical approximation may be the way to go, in this case.

Eliz.

 

[lollipop2046]

ther original eqt i have is 3n^3 > (n+1)^3
if like i said, just solve this algebraically, can i apply log of Ln to both sides?
but then i got n > (1/3)^(1/3) which does not hold for the inequality... did i do sth wrong for applying log and ln?

 

[stapel]

I don't know about the legality (mathematically) of taking logs of values of whose sign you are not certain. And even if you could do that, I don't see how you got the answer you claim, since the "n" would be inside the logs.

Please reply showing all of your steps and reasoning, beginning with the actual exercise in question. Thank you.

Eliz.

 

[lollipop2046]

actually the orignial question is to show that "3^n > n^3"
as i have proved for all cases where n <= 0, all i need is n>0.

if i do it by induction starting 0, i encountered a problem at n=3 since both side of inequality are equal whcih makes my statement false.
(actually can i do induction like this: check for n=1, assume true for n EXCEPT SPECIAL CASE FOR n=k)?

anywys, if i assume true for n, i need to prove n+1.
3^(n+1) >? (n+1)^3; and i know that 3*3n > 3n^3,
so i want to show that 3n^3 > (n+1)^3.....

 

[stapel]

Okay, so this wasn't actually a "solving the polynomial inequality by graphing" problem, but was likely an induction problem. Would it be safe to assume that "n" takes on only integral values?

(Note: The strict inequality is false for n = 3.)

You can show, by computation, that the statement is true for n = 1, 2, and 4. Then assume the statement for n = k > 4. Then use the size of k in the "n = k + 1" step.

. . . . .3^k+1 = 3(3^k)

. . . . .> 3(k³)

. . . . .= k³ + k³ + k³

. . . . .> k³ + 4k² + 16k

...and so forth.

Eliz.