SOLUTION: Suppose the domain of f is (-1,3). Define the function g by g(x) = 5 - f(x) + f(5/x). What is the domain of g?

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Question 1209614: Suppose the domain of f is (-1,3). Define the function g by
g(x) = 5 - f(x) + f(5/x).
What is the domain of g?

Answer by ikleyn(52797)   (Show Source): You can put this solution on YOUR website!
.
Suppose the domain of f is (-1,3). Define the function g by
g(x) = 5 - f(x) + f(5/x).
What is the domain of g?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let D be the domain of f(x),  D = (-1,3).


Both x and 5/x should be in the interval (-1,3),
and, additionally, x should not be equal to 0 (zero).


So, the constraints are

    -1 < x < 3,    (1)

    -1 < 5/x < 3,  (2)

     x =/= 0.      (3)


Inequalities (1) and (3) tell us that we should consider two separate intervals (-1,0) and (0,3) for x,
and determine other limitations on x from inequality (2) 



(a)  So, let x be in the interval (-1,0).  Thus, x is negative now.

     Then first of the two inequalities (2),  -1 < 5/x, is equivalent to

         -x > 5  (after multiplying both sides by negative value of x and flipping the inequality sign),

     or, which is the same,

          x < -5.


          Thus we determined that if x is in (-1,0), then due to first inequality (2), 
                  x must be lesser than -5, which is out of the domain D.

          So, we may exclude this case "x is in (-1,0)"  from our consideration.



(b)  Now, let x be in the interval (0,3).  Thus, x is positive now.

     Then first of the two inequalities (2),  -1 < 5/x, is always valid and does not imply other restrictions on x.

     The second of the two inequalities (2),  5/x < 3,  then implies  x > 5/3.


            Thus, if x is positive, then due to second inequality of (2), 
                        x must be greater than 5/3.


Combining what we found in (a) and (b), the answer to the problem's question is


     +-------------------------------------------------+
     |   The domain of the function g(x) is (5/3,3).   |
     +-------------------------------------------------+

Solved.



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