Math Problem Statement
Solution
Let's analyze the function , which has zeros at , , and . The problem asks for the sign of the function on the interval .
Step-by-step analysis:
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Factoring the function: The function is already factored as .
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Critical points: The zeros (roots) of the function are:
These are the points where the function equals zero.
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Interval of interest: We are interested in the sign of the function on the interval , which lies between two of the critical points and .
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Test points: To determine the sign of the function in this interval, we can pick a test point inside . A good choice is , as it lies between these two values.
Now, evaluate the sign of each factor at :
- (negative)
- (positive)
- (negative)
So, , which is positive.
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Conclusion for the interval: Since the test point yields a positive result, and there are no zeros within this interval except at the boundaries, the sign of the function is consistent on this interval. Therefore, is always positive on the interval .
Answer:
The correct answer is A. is always positive on the interval.
Would you like more details or explanations on any step? Here are some follow-up questions for deeper understanding:
- What is the significance of the function's critical points in determining its behavior?
- How can the sign of a function change across different intervals?
- What other test points could be chosen to verify the function's sign in an interval?
- How would the sign of the function differ in other intervals defined by the critical points?
- Can you explain how multiplicity of roots affects the sign change at a root?
Tip: When analyzing the sign of a polynomial, testing values within different intervals defined by the zeros is a reliable method to check for sign changes.
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Math Problem Analysis
Mathematical Concepts
Polynomial Functions
Factoring
Zeros of a Function
Interval Testing
Formulas
Polynomial Factoring
Sign Testing by Substitution
Theorems
Intermediate Value Theorem
Sign Changes in Polynomials
Suitable Grade Level
Grades 10-12
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