So I have this function
$f(t)=\log(t^2+5t)$
And I want to find the domain.
Now I am taking into consideration $t^2+5$ which I know needs to be greater than $0$ and $t$ that needs to be greater than $0$. And until here im good.
On the answers sheet it says that $t>0$ and that's fine.
But then it says that $t+5>0$ when $t<-5$. And that for me doesn't make any sense whatsover. I take a number that is smaller than $t$ since that's what $t<-5$ implies. E.g $-7$. Then $-7+5$ is NOT a positive number and is not greater than $0$.
I thought that t+5>0 when t>-5. Then the domain would be from (-5,0) and from (0 to inifity).
I have no idea what I am not understanding , I have dyscalulia and any simple help would be great . Thanks!
The answer that the professor told me is (−∞, −5) ∪ (0, +∞).
Best Answer
$$f(t) = \log(t^2+5t)$$
You need the argument of the logarithm to be positive. Hence, you have
$$t^2+5t > 0$$
$$t(t+5) > 0$$
For a product to positive, either both factors are positive or both are negative. (This is an alternative to the other procedure posted.)
Case $1$:
$$t > 0 \quad t+5 > 0$$
$$t > 0 \quad t > -5$$
For $t > 0$, both conditions are true, so the factors are positive.
Case $2$:
$$t < 0 \quad t+5 < 0$$
$$t < 0 \quad t < -5$$
For $t < -5$, both conditions are true, so the factors are negative.
Hence, the domain becomes $(-\infty, -5)\cup(0, +\infty).$