SOLUTION: I am no good at graphing. Please help. Find and graph the solution set: a) x^3 -16x >0 b) X^2-5x_< -6

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Question 87680: I am no good at graphing. Please help.
Find and graph the solution set:
a) x^3 -16x >0
b) X^2-5x_< -6
Found 2 solutions by josmiceli, Edwin McCravy:
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Look at it this way. Spend 10 times as much time thinking before you even
try to graph. You're probably too much in a hurry to get it done.
It's like skiing,skateboarding- you can do better-think more about what you're doing.



Now you start thinking
What would make this inequality NOT true?
One of the things is if the left side turned out to be 0.
If any of the 3 factors on the left is 0, then the left is 0.
Call the left side f(x). So far we have
x - - - f(x)
+4 - - - 0
-4 - - - 0
0 - - - 0
These are the roots of f(x).
What else makes the inequality NOT true?
You have 3 factors. What does it take to make a negative result?
A negative result would make the inequality NOT true
(+)(+)(+) = +
(-)(-)(+) = +
(-)(+)(-) = +
(+)(-)(-) = +
These combinations make the inequality TRUE -not what we want
---------------------------
(+)(+)(-) = -
(+)(-)(+) = -
(-)(+)(+) = -
(-)(-)(-) = -
This thinking helps you make SENSE of the graph and is worth the effort
How can we get any of these last 4 combinations? Try different values
for x and see
What if ? Just look at the factors in terms of sign only
If x is 0, then
(0)(+)(-) > 0 NOT true
---------------------------
If , then
(+)(+)(-) > 0 NOT true
---------------------------
If . then
(-)(+)(-) > 0 This is TRUE
---------------------------
So far, we have if the inequality is NOT true
What about For that I get
(-)(-)(-) > 0 This is NOT true
---------------------------
The only thing remaining is
I get
(+)(+)(+) > 0 and this is TRUE
---------------------------
The end result of all this is that the inequality does NOT hold
for the cases
and

For ALL other cases the inequality holds
Now I draw the graph and prove it (or not, if I'm wrong)

The graph seems to verify my statements.
It's (+) for and also for the
interval
You could reach the same result just by testing a lot of numbers
and seeing if f(x) is (-), (0), or (+). That's what I do as a last
resort.
Hope this helps- you can apply it to your 2nd problem
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!

a) x³ - 16x > 0

Since we already have 0 on the right side, we factor out
x on the left side:

 x(x² - 16) > 0

Then we factor the expression in parentheses as the
difference of two squares:

 x(x - 4)(x + 4) > 0

The critical values of the left side are 0, 4 and -4

So we put these on a number line and mark them with
an open circle.  (Open circle since it's > and not >

--------o--------o--------o-----------
       -4        0        4

Now we substitute any value to the left of -4 to see
if we shade that part or not. Suppose we choose -5.
Substitute x = -5 into:

   x(x - 4)(x + 4) > 0
-5(-5 - 4)(-5 + 4) > 0
        -5(-9)(-1) > 0
               -45 > 0

This is false so we do not shade the part of the number
line left of -4, so the number line is unchanged:

--------o--------o--------o-----------
       -4        0        4

Nex we substitute any value between -4 and 0 to see
if we shade that part or not. Suppose we choose -1.
Substitute x = -1 into:

   x(x - 4)(x + 4) > 0
-1(-1 - 4)(-1 + 4) > 0
         -1(-5)(3) > 0
                15 > 0

This is true so we do shade the part of the number
line between -4 and 0, so the number line is now:

--------o========o--------o-----------
       -4        0        4

Next we substitute any value between 0 and 4 to see
if we shade that part or not. Suppose we choose 1.
Substitute x = 1 into:

   x(x - 4)(x + 4) > 0
   1(1 - 4)(1 + 4) > 0
          1(-3)(5) > 0
               -15 > 0

This is false so we do not shade the part of the number
line between 0 and 4, so the number line is still:

--------o========o--------o-----------
       -4        0        4

Finally we substitute any value to the right of 4 to see
if we shade that part or not. Suppose we choose 5.
Substitute x = 5 into:

   x(x - 4)(x + 4) > 0
   5(5 - 4)(5 + 4) > 0
           5(1)(9) > 0
                45 > 0

This is true so we do shade the part of the number
line right of 4, so the final number line is now:

--------o========o--------o===========>
       -4        0        4

The interval notation is an abbreviation of this graph

     (-4, 0) U (4, oo)

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

b) x² - 5x < -6

Get 0 on the right by adding 6 to both sides:

   x² - 5x + 6 < 0

Factor the left side:

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

The critical values are 2 and 3

So we put these on a number line, this time using
closed circles because the inequality this time is 
< and not <.

------------------l----l-------
  -1    0    1    2    3    4    

Now we substitute any value to the left of 2 to see
if we shade that part or not. Suppose we choose 0.
Substitute x = 0 into:

    (x - 2)(x - 3) < 0
    (0 - 2)(0 - 3) < 0
          (-2)(-3) < 0
                 6 < 0 

This is false so we do not shade the part of the number
line left of 2, so the number line is unchanged:

------------------l----l-------
  -1    0    1    2    3    4    

Now we substitute any value between 2 and 3 to see
if we shade that part or not. Suppose we choose 2.5.
Substitute x = 2.5 into:

    (x - 2)(x - 3) < 0
(2.5 - 2)(2.5 - 3) < 0
         (.5)(-.5) < 0
              -.25 < 0 

This is true so we do shade the part of the number
line between 2 and 3, so the number line is now:

------------------l====l-------
  -1    0    1    2    3    4    

Finally we substitute any value to the right of 3 
to see if we shade that part or not. Suppose we 
choose 4.  Substitute x = 4 into:

    (x - 2)(x - 3) < 0
    (4 - 2)(4 - 3) < 0
            (2)(1) < 0
                 2 < 0 

This is false so we do not shade the part of the number
line to the right of 3, so the final number line is this:

------------------l====l-------
  -1    0    1    2    3    4

   In interval notation that is [2, 3]

Edwin
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