4 x^3 + 12x^2 = 0
4 x^2 (x + 3) = 0
x = −3 or x = 0
We can test where the derivative is positive and negative by picking a point in each of the three
regions −∞ < x < −3, −3 < x < 0, and 0 < x < ∞, plugging the point into the derivative, and
seeing what the sign of the answer is. Because x^2 is never negative, you should find that the
derivative is negative on the interval −∞ < x < −3.
- C We will need to use the fact that = 1 to find the limit.
First, rewrite the limit as
Next, break the expression into two rational expressions.
This can be broken up further into
We will evaluate the limit of each separately.
First expression
Divide the top and bottom by x: .
Then, multiply the top and bottom of the upper expression by 3, and the top and bottom of the
lower expression by 5: .
