# How do you find the domain and range of f(t)=-11/sqrtt?

##### 1 Answer
May 22, 2016

$t \in \left(0 , \infty\right)$, $f \left(t\right) \in \left(- \infty , 0\right)$

#### Explanation:

The domain is the "allowed" values for $t$ (the input of the function) and the range is the values that can be returned by the function given that you use the appropriate values for $t$.

Assuming you are working strictly with the real numbers, $\mathbb{R}$, then:

For the domain:

We can't have $t = 0$ on the bottom otherwise we would be dividing by $0$. Also we can't take the square root of a negative so anything less than or equal to 0 is not allowed. (i.e 0 is the lowest allowed number on the domain). There is no limit on the highest number so the domain is:

$t \in \left(0 , \infty\right)$

For the range:

We have already deduced that the lowest value for $t$ is $0$. We find that as we make $t$ very small the fraction becomes very big, in fact $f \left(t\right)$ gets infinitely big as $t$ approaches zero. (Don't forget the negative). So the lower limit on our range will be $- \infty$. On the contrary, as we make $t$ infinitely big, the larger value on our domain, $f \left(t\right)$ gets very small and approaches $0$ when $t$ gets to infinity. So the other limit on our range is $0$ Hence the range is:

$f \left(t\right) \in \left(- \infty , 0\right)$

Note, we have used curved brackets to indicate an "open interval." This means the limit specified is NOT part of the set. That is: $t$ can be greater than $0$ but not equal to $0$.

Indeed there do exist cases where the limiting numbers are also part of the domain or range, in which case we use a square bracket $\left[x , y\right]$ to indicate a closed interval.