SOLUTION: Can anyone check A for me please. I'm lost on B. For C would I just plug in (t+1) for x? Let f(x)= x^2/(2x-3). (a) Calculate f(-2) = (-2)^2/(2(-2)-3) = 4/(-4-3) = 4/

Algebra ->  Coordinate-system -> SOLUTION: Can anyone check A for me please. I'm lost on B. For C would I just plug in (t+1) for x? Let f(x)= x^2/(2x-3). (a) Calculate f(-2) = (-2)^2/(2(-2)-3) = 4/(-4-3) = 4/      Log On


   

Question 1144720: Can anyone check A for me please. I'm lost on B. For C would I just plug in (t+1) for x?
Let f(x)= x^2/(2x-3).
(a) Calculate f(-2)
= (-2)^2/(2(-2)-3)
= 4/(-4-3)
= 4/-7
= -4/7

(b) State the domain of the function in interval notation.



(c) Find and simplify as much as possible: f(t+1)
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
a) You got it. Way to go!
b) The domain of function is the set of all possible values of the independent variable (x). Here, you can not have 2x-3 = 0 --> +x+%3C%3E+3%2F2+, while any other value for x is valid. In interval notation: x can take on any value in these open intervals: (-infinity+, +3%2F2) and (3%2F2, infinity)
Note: An open interval () does not include its endpoints. To signal that an endpoint IS included you would use square brackets [ ] (half-open intervals are possible for some problems as well).
c) Yes, plug in x = t+1 and then simplfy to the extent possible (hint: not a whole lot of simplification will be possible).