SOLUTION: O.K My instructions were " for each function, find the specified function value, if it exists." The problem is f(t)= square root of 5t-10; f(6), f(2), f(1), f(-1) I can't figur

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Question 30019: O.K My instructions were " for each function, find the specified function value, if it exists." The problem is
f(t)= square root of 5t-10; f(6), f(2), f(1), f(-1)
I can't figure out how to get the answers and what is "Does Not Exist"?
PLease help and thanks
Answer by sdmmadam@yahoo.com(530) (Show Source):
You can put this solution on YOUR website!
f(t)= square root of 5t-10; f(6), f(2), f(1), f(-1)
f(t)= sqrt(5t-10) ----(1)
Since in the real number system the sqrt of a negative quantity is not defined we say that for some t if we happen to get the quantity under the root as a negative one then such a thing is not defined which means we cannot find an image f(t) for this t because such a t cannot be entertained in the domain for whom we are not able to give it an associate in the codomain for the funciton f
When we say that f is a function from the real number system R to R we mean
f is a rule that 1)HAS TO give an associate called image in the second set(codomain set) for EVERY t in the first set (domain set) And 2)no member of the domain can have more than one image in the codomain set.
Putting t=6 in (1)
f(6) =sqrt[5X6-10]=swrt(30-10) = sqrt(20) = sqrt(4X5) =2 times sqrt of 5
Putting t=2 in (1)
f(2) =sqrt[5X2-10]=swrt(10-10) = sqrt(0) = 0
Putting t=1 in (1)
f(1) =sqrt[5X1-10]=swrt(5-10) = sqrt(-5) not defined in the real number system
So f(1) does not exist.
(Got the point!)
Putting t=-1 in (1)
f(-1) =sqrt[5X(-1)-10]=swrt(-5-10) = sqrt(-15) not defined in the real number system
So f(-1) does not exist!.
Answer: f(6) = 2 times sqrt of 5
f(2) = 0
f(1) does not exist.
f(-1) does not exist.