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

1 Answer
May 22, 2016

Answer:

#tin (0,oo)#, #f(t) in (-oo,0)#

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, #RR#, 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:

#tin (0,oo)#

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(t)# gets infinitely big as #t# approaches zero. (Don't forget the negative). So the lower limit on our range will be #-oo#. On the contrary, as we make #t# infinitely big, the larger value on our domain, #f(t)# 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(t) in (-oo,0)#

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 #[x,y]# to indicate a closed interval.