1. ## Basic inequality

Cannot solve this for the life of me...

When is 2t / (t+1)^3 > 0 ?
I seem to think it is (-infinity,-1)U[0,infinity), but then I am unsure.

The above expression is also equivelant to 2 / (t^2 (1 + 1/t)^3) - and that is only positive for t > -1.
I am extremely confused!! Help!!! Thanks!

2. Hello, Volcanicrain!

When is: . $\frac{2t}{(t+1)^3} \:>\: 0$

I seem to think it is: . $(\text{-}\,\infty, \text{-}1) \cup (0,\infty)$ . . . . Right!

The above expression is also equivelant to: . $\frac{2}{t^2\left(1 + \frac{1}{t}\right)^3} \:>\:0$ . . . . I agree
. . and that is only positive for $t > -1$ . . . . no
We have: . $\frac{2t}{(t+1)^3} \;>\;0$

A fraction is positive if the numerator and denonominator are: . $\begin{Bmatrix}(1) & \text{ both positive} \\ (2) & \text{ both negative} \end{Bmatrix}$

(1) Both positive: . $\begin{Bmatrix}
2t \:>\:0 & \Rightarrow & t \:>\:0 \\
(t+1)^3 \:>\:0 & \Rightarrow & t+1 \:>\:0 & \Rightarrow & t \:>\:\text{-}1 \end{Bmatrix} \quad\Rightarrow\quad t \:>\:0$

(2) Both negative: . $\begin{Bmatrix}
2t \:<\:0 & \Rightarrow & t \:<\:0 \\
(t+1)^3 \:<\:0 & \Rightarrow & t+1 \:<\:0 & \Rightarrow & t \:<\:\text{-}1 \end{Bmatrix} \quad\Rightarrow\quad t \:<\:\text{-}1$

Therefore: . $(t \,<\,\text{-}1) \vee (t \,>\,0) \quad\Longleftrightarrow\quad (\text{-}\,\infty, \text{-}\,1) \cup (0, \infty)$

3. Could you explain why it's not positive only for t > -1 in my quote though?? I've thought about it all day and I can't understand why my algebra is wrong.

Thank you for confirming my above work though! I am quite grateful!