Help please!

1.) The solution set of the inequality x+5/x-2<=6 in interval notation

                x + 5
               ------- £ 6
                x - 2 

Get 0 on the right by subtracting 6 from both sides:

                x + 5
               ------- - 6 £ 0
                x - 2

Write 6 over 1

                x + 5     6
               ------- - --- £ 0
                x - 2     1

LCD = x - 2.  The first fraction already has that
LCD.  The make the second fraction have it too, we
multiply top and bottom by it:

                x + 5     6(x - 2)
               ------- - ---------- £ 0
                x - 2     1(x - 2)

                x + 5     6x - 12
               ------- - ---------- £ 0
                x - 2      x - 2

Combine the two fractions:

                x + 5 - (6x - 12)
               ------------------- £ 0
                     x - 2     

                x + 5 - 6x + 12
               ------------------ £ 0
                     x - 2

                -5x + 17
               ---------- £ 0
                  x - 2

The zero of the numerator is 17/5 or 3.4
The zero of the denominator is 2

So the critical values are 2 and 3.4

We draw a number line and mark those two 
values:

-----------o----------o-------------
           2         3.4

Select a test value less than 2, say 0,

Substitute it into

                -5x + 17
               ---------- £ 0
                  x - 2

                -5(0) + 17
               ------------ £ 0
                  (0) - 2

                    -17/2 £ 0 

This is true, so we shade that part of 
the number line


< ===========o----------o-------------
             2         3.4

Now choose a test value between 2 and 3.4,
say 3

Substitute it into

                -5x + 17
               ---------- £ 0
                  x - 2

                -5(3) + 17
               ------------ £ 0
                  (3) - 2

                    2 £ 0 

This is false, so we do not shade that part of 
the number line.

< ===========o----------o-------------
             2         3.4

Now choose a test value greater than 3.4,
say 4

Substitute it into

                -5x + 17
               ---------- £ 0
                  x - 2

                -5(4) + 17
               ------------ £ 0
                  (4) - 2

                  -3/2 £ 0 

This is true, so we shade that part of 
the number line

< ===========o----------o============ >
             2         3.4

Now we have to test the critical points themselves.

x cannot be 2 because that causes the denominator 
to be 0, which is undefined. However x can be 3.4
because that makes only the numerator 0 and since
the inequality is " £ " and not " < ", we can
include 0.  So we darken the circle at 3.4.

< ===========o----------·============ >
             2         3.4

So the interval notation for the solution is

  (-¥, 2) È [3.4, ¥) 


2.) If x=243, then x-1/3 is ?

    Notice that 243 = 3·3·3·3·3 = 35
                                   1
    x-1/3 = 243-1/3 = (35)-1/3 = ----------- =
                                 (35)1/3

        1         1          1            1          1 
    -------- = -------- = --------- = --------- = -------- =
      35/3        31+2/3     31·32/3       3·32/3     3(³Ö3²)

        1     ³Ö3        ³Ö3       ³Ö3     ³Ö3
    --------·----- =  --------- = ----- = ----  
     3(³Ö3²)  ³Ö3      3(³Ö3³)     3·3      9



3.) The inequality -4x+7<=2x+4 is equivalent to ?
    
    Solve it like an equation until the last step:

                  -4x + 7 < 2x + 4
                      - 7      - 7
               -----------------------
                  -4x     < 2x - 3
                  -2x      -2x   
               -----------------------
                  -6x     <     -3

But when you get to the last step, where you divide
by the coefficient of x, if you divide by a POSITIVE
number you KEEP the inequality sign as is, but if
you divide by a NEGATIVE number you must REVERSE the
inequality sign.  Here we must divide by -6, which is
NEGATIVE, so we must change the < to >.

                -6x/(-6)  > -3/(-6

                        x > 1/2
 

4.) What is the largest zero of the function f(x)=x^3+5x^2+5x-2

The only way to find it is to approximate it by using a graphing 
calculator. It is approximately 0.3027756377




Edwin