Question 54409
f(x) must be greater than the value of zero
0 < x(x + 3)(x - 1)
The zeros for x(x + 3)(x - 1) is at x=0, x=-3, and x=1.
Pick Points:
<```(-3)```(0)```(1)```>
x = -4
0 < (-4)(-4 + 3)(-4 - 1)
0 < (-4)(-1)(-5)
0 < -20 Not True
<---(-3)```(0)```(1)```>
x = -1
0 < (-1)(-1 + 3)(-1 - 1)
0 < (-1)(2)(-2)
0 < 4 True
<---(-3)====(0)```(1)```>
x = 0.5
0 < (0.5)(0.5 + 3)(0.5 - 1) We already know we will get a negative number, so this is false.
<---(-3)====(0)----(1)```>
x = 2
0 < (2)(2 + 3)(2 - 1)
0 < (2)(5)(1)
0 < 10 This is true.
<---(-3)====(0)----(1)====>
(-3,0)U(1,+{{{infinity}}})