MODERATE EXERCISES

(1) For each of the following functions f(x), 
find the intervals on which: (i) f(x) > 0; and (ii) f(x) < 0.  (Note: for 
some questions, there may be no such interval.)

 (a) f(x) = (2x - 3)(x + 3)(x + 1)

 (b) f(x) = x(x - 5)(x + 3)

 (c) f(x) = 2/3x(x - 4)2

 (d) f(x) = (x - 2)(x2 + 6x + 8)

 (e) f(x) = 2x2(2x - 4)

 (f) f(x) = x/5(1 - x2)

 (a) f(x) = (2x - 3)(x + 3)(x + 1)

     2x - 3     - - -        - - -         - - -           + + +
     x + 3      - - -        + + +         + + +           + + +
     x + 1      - - -        - - -         + + +           + + +
            _____________o____________o_______________o_____________
                        -3           -1              3/2
     f(x)       - - -        + + +         - - -           + + +

     f(x) > 0  on  (-3,-1)  (3/2,)
     f(x) < 0  on  (-,-3)  (-1,3/2)


 (b) f(x) = x(x - 5)(x + 3)

     x         - - -     - - -             + + +            + + +
     x - 5     - - -     - - -             - - -            + + +
     x + 3     - - -     + + +             + + +            + + +

            __________o____________o______________________________
                     -3            0                    5
     f(x)      - - -     + + +             - - -            + + +

     f(x) > 0  on  (-3,0)  (5,)
     f(x) < 0  on  (-,-3)  (0,5)


 (c) f(x) = (2x/3)(x - 4)2

     Since (x - 4)2 is ALWAYS greater than or equal to 0, just use 2x/3

     2x/3 > 0  when  x > 0  and 2x/3 < 0  when  x < 0

     Therefore f(x) > 0  on  (0,4)  (4,)
     f(x) < 0  on  (-,0)


 (d) f(x) = (x - 2)(x2 + 6x + 8)
          = (x - 2)(x + 4)(x + 2)

     x - 2   - - -       - - -              - - -              + + +
     x + 4   - - -       + + +              + + +              + + +
     x + 2   - - -       - - -              + + +              + + +

            __________o____________o________________________o__________
                     -4           -2                        2
     f(x)    - - -       + + +              - - -              + + +

     f(x) > 0  on  (-4,-2)  (2,)
     f(x) < 0  on  (-,-4)  (-2,2)


 (e) f(x) = 2x2(2x - 4)

     Since 2x2 is ALWAYS greater than or equal to 0 we just use (2x - 4)

     2x - 4 > 0  ==>  x > 2

     2x - 4 < 0  ==>  x < 2

     So f(x) > 0  on  (2,)
        f(x) < 0  on  (-,0)  (0,2)

 (f) f(x) = x/5(1-x2)
          = x/5(1-x)(1+x)    

      x/5      - - -       - - -     + + +        + + +         
      1-x      + + +       + + +     + + +        - - -    
      1+x      - - -       + + +     + + +        + + +

             ----------o----------o-----------o-------------
                      -1          0          +1
     f(x)      + + +       - - -      + + +        - - -


     f(x) > 0  on  (-,-1)  (0,1)
     f(x) < 0  on  (-1,0)  (1,)