2(x^3-2x^2+3)
How can I solve this?? Please help!
How can I solve this?? Please help!
-
2x^3 - 4x^2 + 6 < x^3 - x
Subtract x^3 - x from both sides:
x^3 - 4x^2 + x + 6 < 0
For x=0, this is false. For large positive x, the x^3 term dominates, so it's also false. By Descartes' Rule of Signs, this can have at most one positive root. But since f(0) and f(100) are both positive, there would have to be TWO positive roots for this function to be negative for any positive x.
Now let's look at negative x. For large negative x, the x^3 term dominates. There must be at least one negative root. And by Descartes' Rule of Signs, there can't be more than one negative root. Therefore there's one negative root and two complex roots. So it all comes down to finding the negative root of this cubic:
f(x) = x^3 - 4x^2 + x + 6
And are we lucky! f(0) = +6, but f(-1) = 0
Your answer is x < -1.
Nothing else to say.
Subtract x^3 - x from both sides:
x^3 - 4x^2 + x + 6 < 0
For x=0, this is false. For large positive x, the x^3 term dominates, so it's also false. By Descartes' Rule of Signs, this can have at most one positive root. But since f(0) and f(100) are both positive, there would have to be TWO positive roots for this function to be negative for any positive x.
Now let's look at negative x. For large negative x, the x^3 term dominates. There must be at least one negative root. And by Descartes' Rule of Signs, there can't be more than one negative root. Therefore there's one negative root and two complex roots. So it all comes down to finding the negative root of this cubic:
f(x) = x^3 - 4x^2 + x + 6
And are we lucky! f(0) = +6, but f(-1) = 0
Your answer is x < -1.
Nothing else to say.