Expert: Paul Klarreich - 2/26/2009

Question
Consider this function [3x-1/2x) on restricted domain D={x|x<0}. Find the range of this function on the restricted domain.

==============================================================

Is the range y∈R|y≠0

Answer
Questioner:   Heneri
Category:  Calculus
Private:  No
 
Subject:  Function inverse ????

Question:  Consider this function [3x-1/2x) on restricted domain D={x|x<0}. Find the range of this function on the restricted domain.

.................
Is this what you mean?
      3x - 1
f(x) = ------, for x < 0
        2x


==============================================================

Is the range y∈R|y≠0

I don't think so.  Let's work it out a bit:

First, a quick and dirty graph suggests that the graph lies above  y = 3/2.  In fact, some work confirms this:
             1
f(x) = 3/2 - ---,  for  x < 0.
             2x

Now if x < 0 (read:  "x is negative") then -1/2x is positive, so we are right -- the graph lies above y = 3/2.

So I think that is your range:  { y | y > 3/2 }

Still not sure?  How about turning points?

f'(x) = + 1/(2x^2).  There are no turning points.