For what values of t is 8t^3-27t^2+30t+11 less than zero?

Answer 1

f(t) = 8t^3-27t^2+30t+11
f '(t) = 24 t^2 - 54t +30 =0

12 t^2 -27t +15 =0

Discriminant = 27^2 -4(15)(12) = 9

t1 = ( 27 -3)/24 = 1
t2 = (27 +3)/24 = 30/24 = 5/4

f(1) = 8 -27 +30 +11 = 22
f(5/4) = f(1.25) = 14.4375

F(t) increases from minus infinity to 22, then decreases from 22 to 14.4375, then increases from 14.4375 to plus infinity.

we see that the function has negative values only in the domain ] - infinity ; 1[
The only way to find exactly where the function is negative is to find a root in that domain. This can be done either by analytic methods (solving for cubic equations) or graphing the function.

I found a numerical solution
x = -0.2865136322

Hence the function is less than zero for every t in the domain ] - infinity; -0.2865136322 [

Good luck !

Answer 2

let y = 8t^3-27t^2+30t+11
at first, I can not get your question
the graph is thin w/r to y
x = - 0.286514 . .. all values left to this x will have value of y below zero

Answer 3

t < - 0.286