f(x)=(3x-1).(x+2)
3x-1=0⇔x=\(\dfrac{1}{3}\)
x+2=0⇔x=-2
BXD:
x | -∞ -2 \(\dfrac{1}{3}\) +∞ |
3x-1 | - - 0 + |
x+2 | - 0 + + |
f(x) | + 0 - 0 + |
vậy f(x)=0 với ∀ x ∈ {-2;\(\dfrac{1}{3}\)}
f(x)>0 với ∀ x ∈ (-∞;-2) U (\(\dfrac{1}{3}\);+∞)
f(x)<0 với ∀ x ∈ (-2;\(\dfrac{1}{3}\))