Find the positive and negative domains of the following function:
y=(31x+61)(−x−451)
Then determine for which values of x the following is true:
f(x) < 0
The function y=(31x+61)(−x−451) requires us to analyze the sign of the product for various x values.
First, we must find the zeros of each factor:
- The zero of 31x+61 is found by solving 31x+61=0:
Subtract 61 to get:
31x=−61
Multiply both sides by 3:
x=−21.
- The zero of −x−451 is found by solving −x−451=0:
Add 451 to get:
−x=451
Multiply by −1 to find:
x=−451.
Next, we identify the intervals defined by these zeros: x<−451, −451<x<−21, and x>−21.
We will determine the sign of the function in each interval:
- In x<−451:
- 31x+61 and −x−451 are both negative (since both points are below their respective roots), resulting in a positive product.
- In −451<x<−21:
- 31x+61 is negative and −x−451 is positive, resulting in a negative product.
- In x>−21:
- 31x+61 and −x−451 are both positive, resulting in a positive product.
The function is negative in the interval −451<x<−21. Thus, the correct answer corresponding to where the function is negative is the complementary intervals x>−21 or x<−451, which matches choice 2.
Therefore, the solution is x>−21 or x<−451.
x > -\frac{1}{2} or x < -4\frac{1}{5}