SOLUTION: <pre><b> I am confused on these type of problems, please help. If f(x)=x(x+3)(x-1), use interval notation to give all values of x where f(x)>0 a. (-3,1) b. (-3,0)<font fac

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: <pre><b> I am confused on these type of problems, please help. If f(x)=x(x+3)(x-1), use interval notation to give all values of x where f(x)>0 a. (-3,1) b. (-3,0)<font fac      Log On


   

Question 78478: 

I am confused on these type of problems, please help. 
If f(x)=x(x+3)(x-1), use interval notation to give all values of
   x where f(x)>0
a. (-3,1)
b. (-3,0)È(1,¥)
c. (1,3)
d. (0,1)È(3,¥)
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

I am confused on these type of problems, please help. 
If f(x)=x(x+3)(x-1), use interval notation to give all values of
   x where f(x)>0
a. (-3,1)
b. (-3,0)È(1,¥)
c. (1,3)
d. (0,1)È(3,¥)

-------------------------------------

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

The critical values are 0, -3 and 1

Mark those on a number line:

---------o-----------o---o--------
-5  -4  -3  -2  -1   0   1   2   3 

That divides the number line into 4 parts:

 (-¥,-3), (-3,0), (0,1), (1,¥) 

Make this chart:

     interval | (-¥,-3) | (-3,0) | (0,1) | (1,¥) |  
test value, t |    -4   |   -1   |   .5  |    2  |
         f(t) |   -20   |    4   | -.875 |   10  |
 sign of f(t) |     -   |    +   |    -  |    +  |


Since we are looking for the solution set of f(x) > 0
we are looking for the intervals in which the sign of f(x)
is positive. These are (-3,0) and (1,0), so the solution set
is

                  (-3,0) È (1,¥)

which is choice b.

Edwin