ADVANCED EXERCISES

(1) Solve each inequality.

 (a) (x + 5)2(x + 3)(x - 1) > 0

 (b) x(1 - x2)3 > 7(1 - x2)3

 (c) 4x4 - 25x2 + 36  = 0

 (d) 1     1       2
     - + ----- < -----
     x   x + 1   x + 2

 (e) 

 (a) (x + 5)2(x + 3)(x - 1) > 0 
     Since we know that (x + 5)2  0 we just have to worry about
where (x + 5)2 = 0
       x + 5 = 0
       x = -5

     Now we do the factors (x + 3) and (x - 1) either they are both 
     positive or they are both negative

     Assume that they are positive:
     x + 3 > 0   and   x - 1 > 0
     x > -3            x > 1
     Therefore x > 1

     Now assume that they are negitive:
     x + 3 < 0   and   x - 1 < 0
     x < -3            x < 1
     Therefore x < -3

     The answer is (-,-5)  (-5,-3)  (1,)


 (b) x(1 - x2)3 > 7(1 - x2)3

     x(1 - x2)3 - 7(1 - x2)3 > 0

     (x - 7)(1 - x2)3 > 0 

     Note that 1 - x2 has the same sign as (1 - x2)3

     x - 7     - - -     - - -           - - -               + + + 
     1 - x2    - - -     + + +           - - -               - - - 
             __________o________o________________________o__________
                      -1        1                        7
 (x - 7)(1 - x2)3
               + + +     - - -           + + +               - - -

     Therefore the solution is (-,1)  (1,7).


 (c) 4x4 - 25x2 + 36  0

     (x2 - 4)(4x2 - 9)  0

     (x - 2)(x + 2)(2x - 3)(2x + 3)  0

     x - 2  - - -    - - -       - - -         - - -    + + +
     x + 2  - - -    + + +       + + +         + + +    + + +
    2x - 3  - - -    - - -       - - -         + + +    + + + 
    2x + 3  - - -    - - -       + + +         + + +    + + + 
           _________o____o_____________________o____o_________
                   -2   -3/2                  3/2   2
    (x-2)(x+2)(2x-3)(2x+3)
            + + +    - - -       + + +         - - -    + + + 

    Hence the solution is [-2,-3/2]  [3/2,2]


 (d) 1    1     2
     - + --- < ---
     x   x+1   x+2

     2x + 1    2
     ------ < ---
     x(x+1)   x+2

     2x + 1    2
     ------ - --- < 0
     x(x+1)   x+2

     (2x + 1)(x+2) - 2x(x+1)
     ----------------------- < 0
           x(x+1)(x+2)

     2x2 + 5x + 2 - 2x2 - 2x
     ______________________ < 0
           x(x+1)(x+2)

       3x + 2
     ----------- < 0
     x(x+1)(x+2)

     x    - - -       - - -      - - -    - -      + + + 
     x+1  - - -       - - -      + + +    + +      + + +
     x+2  - - -       + + +      + + +    + +      + + +
    3x+2  - - -       - - -      - - -    + +      + + +
         __________o____________o________o____o__________
                  -2           -1      -2/3   0
     3x+2
  -----------  +++    - - -      + + +    - -      + + +
  x(x+1)(x+2)

    The solution is (-2,-1)  (-2/3,0)


 (e) 

     Since we have  in the denominator, we must have x > 0

     (1 - x)2  4x(x - 1)

     (x - 1)2 - 4x(x - 1)  0

     (x - 1)(x - 1 - 4x)  0

     (x - 1)(-3x - 1)  0

     Either both factors are positive or both are negative.
     Let   x - 1  0    and    -3x - 1  0
           x  1               -1  3x
                               -1/3  x
         impossible situation.
     Let   x - 1  0    and    -3x -1  0
           x  1               -1/3  x

     But from the first statement, x > 0.
     Therefore the solution is (0,1]