Answer:
f(x) > 0 over the interval
Step-by-step explanation:
If f(x) is a continuous function, and that all the critical points of behavior change are described by the given information, then we can say that the function crossed the x axis to reach a minimum value of -12 at the point x=-2.5, then as x increases it ascends to a maximum value of -3 for x = 0 (which is also its y-axis crossing) and therefore probably a local maximum.
Then the function was above the x axis (larger than zero) from , until it crossed the x axis (becoming then negative) at the point x = -4. So the function was positive (larger than zero) in such interval.
There is no such type of unique assertion regarding the positive or negative value of the function when one extends the interval from to -3, since between the values -4 and -3 the function adopts negative values.
First we must convert these mixed numbers to improper fractions:
So then we need the product of these two numbers. That means to multiply:
But then we need the value of half of this product. So then let's multiply by one-half:
Then we can convert this from an improper fraction to a mixed number:
And so this number is half of the product of the initial two numbers.